How do I retrieve a list of active items from an Array?
1
I'm trying to get a list of employees who only have active hobbies and whose role is of type 'A'.
I tried using the below query but without any success.
What am I doing wrong?
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>Update
When I added more employees with new hobbies to the array, the accepted answer fails to produce the correct output. Why does this happen?
var people = [
key: 'Employees',
employees: [
name: 'Steve',
age: 50,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
,
active: false,
name: 'skating',
roles: ['C', 'A']
,
active: true,
name: 'snooker',
roles: ['C', 'A']
,
active: true,
name: 'darts',
roles: ['C', 'A']
]
,
name: 'joe',
age: 20,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
active: true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
var employees = people.find(item => item.key === 'Employees').employees.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
javascript arrays dictionary filter
asked Nov 13 '18 at 19:44
QCarQCar
1691033
add a comment |
1
I'm trying to get a list of employees who only have active hobbies and whose role is of type 'A'.
I tried using the below query but without any success.
What am I doing wrong?
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>Update
When I added more employees with new hobbies to the array, the accepted answer fails to produce the correct output. Why does this happen?
var people = [
key: 'Employees',
employees: [
name: 'Steve',
age: 50,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
,
active: false,
name: 'skating',
roles: ['C', 'A']
,
active: true,
name: 'snooker',
roles: ['C', 'A']
,
active: true,
name: 'darts',
roles: ['C', 'A']
]
,
name: 'joe',
age: 20,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
active: true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
var employees = people.find(item => item.key === 'Employees').employees.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
javascript arrays dictionary filter
asked Nov 13 '18 at 19:44
QCarQCar
1691033
1
Just to clarify: "who only have active hobbies" means that all their hobbies have to be active? or just at least one? And "whose role is of type A" means they got at leadt one role? or their active hobby has to have that role? or all their hobbies have to have that role?
– Jonas Wilms
Nov 13 '18 at 19:56
add a comment |
1
1
1
I'm trying to get a list of employees who only have active hobbies and whose role is of type 'A'.
I tried using the below query but without any success.
What am I doing wrong?
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>Update
When I added more employees with new hobbies to the array, the accepted answer fails to produce the correct output. Why does this happen?
var people = [
key: 'Employees',
employees: [
name: 'Steve',
age: 50,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
,
active: false,
name: 'skating',
roles: ['C', 'A']
,
active: true,
name: 'snooker',
roles: ['C', 'A']
,
active: true,
name: 'darts',
roles: ['C', 'A']
]
,
name: 'joe',
age: 20,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
active: true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
var employees = people.find(item => item.key === 'Employees').employees.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
javascript arrays dictionary filter
asked Nov 13 '18 at 19:44
QCarQCar
1691033
I'm trying to get a list of employees who only have active hobbies and whose role is of type 'A'.
I tried using the below query but without any success.
What am I doing wrong?
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>Update
When I added more employees with new hobbies to the array, the accepted answer fails to produce the correct output. Why does this happen?
var people = [
key: 'Employees',
employees: [
name: 'Steve',
age: 50,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
,
active: false,
name: 'skating',
roles: ['C', 'A']
,
active: true,
name: 'snooker',
roles: ['C', 'A']
,
active: true,
name: 'darts',
roles: ['C', 'A']
]
,
name: 'joe',
age: 20,
hobbies: [
active: true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
active: true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
var employees = people.find(item => item.key === 'Employees').employees.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>javascript arrays dictionary filter
javascript arrays dictionary filter
asked Nov 13 '18 at 19:44
QCarQCar
1691033
asked Nov 13 '18 at 19:44
QCarQCar
1691033
edited Nov 13 '18 at 21:14
QCar
asked Nov 13 '18 at 19:44
QCarQCar
1691033
asked Nov 13 '18 at 19:44
QCarQCar
1691033
asked Nov 13 '18 at 19:44
QCarQCar
1691033
1691033
1
Just to clarify: "who only have active hobbies" means that all their hobbies have to be active? or just at least one? And "whose role is of type A" means they got at leadt one role? or their active hobby has to have that role? or all their hobbies have to have that role?
– Jonas Wilms
Nov 13 '18 at 19:56
add a comment |
1
Just to clarify: "who only have active hobbies" means that all their hobbies have to be active? or just at least one? And "whose role is of type A" means they got at leadt one role? or their active hobby has to have that role? or all their hobbies have to have that role?
– Jonas Wilms
Nov 13 '18 at 19:56
1
1
Just to clarify: "who only have active hobbies" means that all their hobbies have to be active? or just at least one? And "whose role is of type A" means they got at leadt one role? or their active hobby has to have that role? or all their hobbies have to have that role?
– Jonas Wilms
Nov 13 '18 at 19:56
Just to clarify: "who only have active hobbies" means that all their hobbies have to be active? or just at least one? And "whose role is of type A" means they got at leadt one role? or their active hobby has to have that role? or all their hobbies have to have that role?
– Jonas Wilms
Nov 13 '18 at 19:56
add a comment |
5 Answers
5
active
oldest
votes
1
Things go wrong where you map the employees to their hobbies: this will make your final result consist of hobbies, not employees.
You need to stick to the employee level:
var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);In an expression there is no need to compare a boolean property with true. Just use the property (active in this case).
Use .some instead of .every if the requirement is that employees have at least one such hobby, instead of requiring that all their hobbies comply with the condition.
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
1
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
1
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose.everyor.somedepending on your expectations.
– trincot
Nov 13 '18 at 21:50
add a comment |
1
.map(emp => emp.hobbies) returns an array of the hobbies, so the value of employees will be the filtered list of hobbies, not the employees that have those hobbies. You need to filter the employees, not map them.
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
add a comment |
1
You can simply do one call to Array.prototype.filter which checks the conditions you mentioned:
person.hobbies.every(y => y.active) && person.hobbies.every(z => z.roles.includes('A'))
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
add a comment |
1
If you do this:
.map(emp => emp.hobbies).filter(hobby =>
you map every employee to its hobbies, which fill result in a 2D array:
[[ active: true , active: false ], [/*...*/]]
Therefore hobby is not one hobby but an array of hobbies.
You said
I'm trying to get a list of employees
... which means you actually don't want to .map to the hobbies, but rather .filter the employees and check if ever hobby fullfills certain rules:
const employees = people.find(( key ) => key === "Employees").employees;
const isActive = hobby => hobby.active && hobby.roles.includes("A");
const result = employees.filter(emp => emp.hobbies.every(isActive));
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))
– QCar
Nov 13 '18 at 22:46
add a comment |
0
You can use map & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)you could also use reduce & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Things go wrong where you map the employees to their hobbies: this will make your final result consist of hobbies, not employees.
You need to stick to the employee level:
var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);In an expression there is no need to compare a boolean property with true. Just use the property (active in this case).
Use .some instead of .every if the requirement is that employees have at least one such hobby, instead of requiring that all their hobbies comply with the condition.
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
1
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
1
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose.everyor.somedepending on your expectations.
– trincot
Nov 13 '18 at 21:50
add a comment |
1
Things go wrong where you map the employees to their hobbies: this will make your final result consist of hobbies, not employees.
You need to stick to the employee level:
var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);In an expression there is no need to compare a boolean property with true. Just use the property (active in this case).
Use .some instead of .every if the requirement is that employees have at least one such hobby, instead of requiring that all their hobbies comply with the condition.
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
1
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
1
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose.everyor.somedepending on your expectations.
– trincot
Nov 13 '18 at 21:50
add a comment |
1
1
1
Things go wrong where you map the employees to their hobbies: this will make your final result consist of hobbies, not employees.
You need to stick to the employee level:
var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);In an expression there is no need to compare a boolean property with true. Just use the property (active in this case).
Use .some instead of .every if the requirement is that employees have at least one such hobby, instead of requiring that all their hobbies comply with the condition.
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
Things go wrong where you map the employees to their hobbies: this will make your final result consist of hobbies, not employees.
You need to stick to the employee level:
var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);In an expression there is no need to compare a boolean property with true. Just use the property (active in this case).
Use .some instead of .every if the requirement is that employees have at least one such hobby, instead of requiring that all their hobbies comply with the condition.
var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);var people = [key: 'Employees',employees: [ name: 'joe', age: 20, hobbies: ['active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: ['active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: ['active': true, name: 'running', roles: ['C'], name: 'Chess', active: false, roles: ['C','A']] ]];
var employees = people.find(item => item.key === 'Employees').employees
.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
edited Nov 13 '18 at 20:06
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
answered Nov 13 '18 at 19:54
trincottrincot
121k1586118
121k1586118
1
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
1
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose.everyor.somedepending on your expectations.
– trincot
Nov 13 '18 at 21:50
add a comment |
1
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
1
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose.everyor.somedepending on your expectations.
– trincot
Nov 13 '18 at 21:50
1
1
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You've given a very detailed answer and have answered my question. Thanks
– QCar
Nov 13 '18 at 20:04
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
You're welcome ;-)
– trincot
Nov 13 '18 at 20:06
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
I just updated my question again, it seems that the code you submitted does not produce the desired result when I add more items to the array. Please see above edit. Thanks
– QCar
Nov 13 '18 at 21:04
1
1
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose
.every or .some depending on your expectations.– trincot
Nov 13 '18 at 21:50
For that data 3 employees are returned. Kate is not among them because not every hobby she has is marked active. I can only repeat the last paragraph of my answer, since your question left ambiguous which of the two logics you were looking for. See also the comment made to your question by Jonas which you did not answer. I believe my answer is correct. You just have to choose
.every or .some depending on your expectations.– trincot
Nov 13 '18 at 21:50
add a comment |
1
.map(emp => emp.hobbies) returns an array of the hobbies, so the value of employees will be the filtered list of hobbies, not the employees that have those hobbies. You need to filter the employees, not map them.
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
add a comment |
1
.map(emp => emp.hobbies) returns an array of the hobbies, so the value of employees will be the filtered list of hobbies, not the employees that have those hobbies. You need to filter the employees, not map them.
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
add a comment |
1
1
1
.map(emp => emp.hobbies) returns an array of the hobbies, so the value of employees will be the filtered list of hobbies, not the employees that have those hobbies. You need to filter the employees, not map them.
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
.map(emp => emp.hobbies) returns an array of the hobbies, so the value of employees will be the filtered list of hobbies, not the employees that have those hobbies. You need to filter the employees, not map them.
// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
emp.hobbies.every(h => h.active && h.roles.includes('A')));
console.log(employees);<script>
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
</script>answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
answered Nov 13 '18 at 19:54
BarmarBarmar
424k35248349
424k35248349
add a comment |
add a comment |
1
You can simply do one call to Array.prototype.filter which checks the conditions you mentioned:
person.hobbies.every(y => y.active) && person.hobbies.every(z => z.roles.includes('A'))
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
add a comment |
1
You can simply do one call to Array.prototype.filter which checks the conditions you mentioned:
person.hobbies.every(y => y.active) && person.hobbies.every(z => z.roles.includes('A'))
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
add a comment |
1
1
1
You can simply do one call to Array.prototype.filter which checks the conditions you mentioned:
person.hobbies.every(y => y.active) && person.hobbies.every(z => z.roles.includes('A'))
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
You can simply do one call to Array.prototype.filter which checks the conditions you mentioned:
person.hobbies.every(y => y.active) && person.hobbies.every(z => z.roles.includes('A'))
var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);var people = [
key: 'Employees',
employees: [
name: 'joe',
age: 20,
hobbies: [
'active': true,
name: 'skating',
roles: ['C', 'A']
]
,
name: 'amy',
age: 32,
hobbies: [
'active': true,
name: 'surfing',
roles: ['A']
]
,
name: 'kate',
age: 34,
hobbies: [
'active': true,
name: 'running',
roles: ['C']
,
name: 'Chess',
active: false,
roles: ['C', 'A']
]
]
];
let employees = people[0].employees.filter(x =>
x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)
console.log(employees);answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
answered Nov 13 '18 at 19:56
connexoconnexo
21.6k73556
21.6k73556
add a comment |
add a comment |
1
If you do this:
.map(emp => emp.hobbies).filter(hobby =>
you map every employee to its hobbies, which fill result in a 2D array:
[[ active: true , active: false ], [/*...*/]]
Therefore hobby is not one hobby but an array of hobbies.
You said
I'm trying to get a list of employees
... which means you actually don't want to .map to the hobbies, but rather .filter the employees and check if ever hobby fullfills certain rules:
const employees = people.find(( key ) => key === "Employees").employees;
const isActive = hobby => hobby.active && hobby.roles.includes("A");
const result = employees.filter(emp => emp.hobbies.every(isActive));
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))
– QCar
Nov 13 '18 at 22:46
add a comment |
1
If you do this:
.map(emp => emp.hobbies).filter(hobby =>
you map every employee to its hobbies, which fill result in a 2D array:
[[ active: true , active: false ], [/*...*/]]
Therefore hobby is not one hobby but an array of hobbies.
You said
I'm trying to get a list of employees
... which means you actually don't want to .map to the hobbies, but rather .filter the employees and check if ever hobby fullfills certain rules:
const employees = people.find(( key ) => key === "Employees").employees;
const isActive = hobby => hobby.active && hobby.roles.includes("A");
const result = employees.filter(emp => emp.hobbies.every(isActive));
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))
– QCar
Nov 13 '18 at 22:46
add a comment |
1
1
1
If you do this:
.map(emp => emp.hobbies).filter(hobby =>
you map every employee to its hobbies, which fill result in a 2D array:
[[ active: true , active: false ], [/*...*/]]
Therefore hobby is not one hobby but an array of hobbies.
You said
I'm trying to get a list of employees
... which means you actually don't want to .map to the hobbies, but rather .filter the employees and check if ever hobby fullfills certain rules:
const employees = people.find(( key ) => key === "Employees").employees;
const isActive = hobby => hobby.active && hobby.roles.includes("A");
const result = employees.filter(emp => emp.hobbies.every(isActive));
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
If you do this:
.map(emp => emp.hobbies).filter(hobby =>
you map every employee to its hobbies, which fill result in a 2D array:
[[ active: true , active: false ], [/*...*/]]
Therefore hobby is not one hobby but an array of hobbies.
You said
I'm trying to get a list of employees
... which means you actually don't want to .map to the hobbies, but rather .filter the employees and check if ever hobby fullfills certain rules:
const employees = people.find(( key ) => key === "Employees").employees;
const isActive = hobby => hobby.active && hobby.roles.includes("A");
const result = employees.filter(emp => emp.hobbies.every(isActive));
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
edited Nov 13 '18 at 19:58
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
answered Nov 13 '18 at 19:52
Jonas WilmsJonas Wilms
56.9k42851
56.9k42851
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))
– QCar
Nov 13 '18 at 22:46
add a comment |
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))
– QCar
Nov 13 '18 at 22:46
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :
var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))– QCar
Nov 13 '18 at 22:46
this is exactly what i want but the query isn't working as expected. I only want to display a list of employees with Active hobbies and who has a role of 'A'. I tried running this query but it does not work :
var result = people.find(item => item.key === 'Employees').employees.filter(emp => emp.hobbies.every(h => h.active && h.roles.includes('A')))– QCar
Nov 13 '18 at 22:46
add a comment |
0
You can use map & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)you could also use reduce & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
add a comment |
0
You can use map & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)you could also use reduce & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
add a comment |
0
0
0
You can use map & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)you could also use reduce & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
You can use map & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)you could also use reduce & filter:
var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees')
.map((employees) =>
employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))
console.log(result)var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)var people = [ key: 'Employees', employees: [ name: 'joe', age: 20, hobbies: [ 'active': true, name: 'skating', roles: ['C', 'A'] ] , name: 'amy', age: 32, hobbies: [ 'active': true, name: 'surfing', roles: ['A'] ] , name: 'kate', age: 34, hobbies: [ 'active': true, name: 'running', roles: ['C'] , name: 'Chess', active: false, roles: ['C', 'A'] ] ] ];
const result = people.filter(x => x.key == 'Employees').reduce((r,employees) =>
r.push(employees.filter(x =>
x.hobbies.some(y => y.active && y.roles.includes('A'))))
return r
, )
console.log(result)answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
edited Nov 13 '18 at 20:03
answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
answered Nov 13 '18 at 19:54
AkrionAkrion
9,41711224
9,41711224
add a comment |
add a comment |
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1
Just to clarify: "who only have active hobbies" means that all their hobbies have to be active? or just at least one? And "whose role is of type A" means they got at leadt one role? or their active hobby has to have that role? or all their hobbies have to have that role?
– Jonas Wilms
Nov 13 '18 at 19:56