How find a value within an array, with a users input
var x1 ='&spades',
y1 ='&clubs'
z1 ='&hearts';
var x2 = ' ', y2 = ' ', z2 = ' ';
var x3 = ' ', y3 = ' ', z3 = ' ';
var Array = [x1,y1,z1,
x2,y2,z2,
x3,y3,z3];
Then I get an input from the user asking them to select a location (ex. x1
) store it as input 1, then ask them for another location (ex. x2
) and store that as input 2, and I have to swap the locations of the value.
I initially tried something like Array[input1] = Array[input2]
But Array[input1]
is undefined even though input1 = x1
because Array[x1]
is still undefined.
So how do I make Array[x1] = Array[0]
so I can swap the values or is there an easier way?
Sorry if my question is poorly formed.
javascript html arrays
|
show 4 more comments
var x1 ='&spades',
y1 ='&clubs'
z1 ='&hearts';
var x2 = ' ', y2 = ' ', z2 = ' ';
var x3 = ' ', y3 = ' ', z3 = ' ';
var Array = [x1,y1,z1,
x2,y2,z2,
x3,y3,z3];
Then I get an input from the user asking them to select a location (ex. x1
) store it as input 1, then ask them for another location (ex. x2
) and store that as input 2, and I have to swap the locations of the value.
I initially tried something like Array[input1] = Array[input2]
But Array[input1]
is undefined even though input1 = x1
because Array[x1]
is still undefined.
So how do I make Array[x1] = Array[0]
so I can swap the values or is there an easier way?
Sorry if my question is poorly formed.
javascript html arrays
For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22
use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25
@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28
As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30
@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30
|
show 4 more comments
var x1 ='&spades',
y1 ='&clubs'
z1 ='&hearts';
var x2 = ' ', y2 = ' ', z2 = ' ';
var x3 = ' ', y3 = ' ', z3 = ' ';
var Array = [x1,y1,z1,
x2,y2,z2,
x3,y3,z3];
Then I get an input from the user asking them to select a location (ex. x1
) store it as input 1, then ask them for another location (ex. x2
) and store that as input 2, and I have to swap the locations of the value.
I initially tried something like Array[input1] = Array[input2]
But Array[input1]
is undefined even though input1 = x1
because Array[x1]
is still undefined.
So how do I make Array[x1] = Array[0]
so I can swap the values or is there an easier way?
Sorry if my question is poorly formed.
javascript html arrays
var x1 ='&spades',
y1 ='&clubs'
z1 ='&hearts';
var x2 = ' ', y2 = ' ', z2 = ' ';
var x3 = ' ', y3 = ' ', z3 = ' ';
var Array = [x1,y1,z1,
x2,y2,z2,
x3,y3,z3];
Then I get an input from the user asking them to select a location (ex. x1
) store it as input 1, then ask them for another location (ex. x2
) and store that as input 2, and I have to swap the locations of the value.
I initially tried something like Array[input1] = Array[input2]
But Array[input1]
is undefined even though input1 = x1
because Array[x1]
is still undefined.
So how do I make Array[x1] = Array[0]
so I can swap the values or is there an easier way?
Sorry if my question is poorly formed.
javascript html arrays
javascript html arrays
edited Nov 12 at 5:46
Andreas
1,7881618
1,7881618
asked Nov 12 at 4:18
David
204
204
For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22
use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25
@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28
As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30
@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30
|
show 4 more comments
For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22
use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25
@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28
As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30
@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30
For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22
For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22
use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25
use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25
@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28
@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28
As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30
As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30
@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30
@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30
|
show 4 more comments
2 Answers
2
active
oldest
votes
Assuming the inputs are strings ('x1'
and 'x2'
), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label
and a value
property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.
Also, better not to assign to a variable named Array
, because that will shadow the global Array
object - name it something else instead, like arr
:
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
If you find that confusing, an alternative is to actually use an intermediate variable:
const orig1 = arr[index1];
arr[index1] = arr[index2];
arr[index2] = orig1;
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
add a comment |
Here's what your code might look like if you use an object:
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assuming the inputs are strings ('x1'
and 'x2'
), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label
and a value
property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.
Also, better not to assign to a variable named Array
, because that will shadow the global Array
object - name it something else instead, like arr
:
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
If you find that confusing, an alternative is to actually use an intermediate variable:
const orig1 = arr[index1];
arr[index1] = arr[index2];
arr[index2] = orig1;
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
add a comment |
Assuming the inputs are strings ('x1'
and 'x2'
), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label
and a value
property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.
Also, better not to assign to a variable named Array
, because that will shadow the global Array
object - name it something else instead, like arr
:
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
If you find that confusing, an alternative is to actually use an intermediate variable:
const orig1 = arr[index1];
arr[index1] = arr[index2];
arr[index2] = orig1;
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
add a comment |
Assuming the inputs are strings ('x1'
and 'x2'
), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label
and a value
property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.
Also, better not to assign to a variable named Array
, because that will shadow the global Array
object - name it something else instead, like arr
:
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
If you find that confusing, an alternative is to actually use an intermediate variable:
const orig1 = arr[index1];
arr[index1] = arr[index2];
arr[index2] = orig1;
Assuming the inputs are strings ('x1'
and 'x2'
), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label
and a value
property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.
Also, better not to assign to a variable named Array
, because that will shadow the global Array
object - name it something else instead, like arr
:
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
If you find that confusing, an alternative is to actually use an intermediate variable:
const orig1 = arr[index1];
arr[index1] = arr[index2];
arr[index2] = orig1;
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
const arr = [
label: 'x1', value: '&spades' ,
label: 'y1', value: '&clubs' ,
label: 'z1', value: '&hearts'
// ...
];
// Inputs:
const label1 = 'x1';
const label2 = 'y1';
// Calculate indicies:
const [index1, index2] = [label1, label2].map(
findLabel => arr.findIndex(( label ) => label === findLabel)
);
([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);
console.log(arr);
answered Nov 12 at 4:25
CertainPerformance
73.3k143454
73.3k143454
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
add a comment |
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52
add a comment |
Here's what your code might look like if you use an object:
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
add a comment |
Here's what your code might look like if you use an object:
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
add a comment |
Here's what your code might look like if you use an object:
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
Here's what your code might look like if you use an object:
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
let arr = ;
arr['x1'] = '&spades';
arr['y1'] = '&clubs';
arr['z1'] = '&hearts';
arr['x2'] = ' ';
arr['y2'] = ' ';
arr['z2'] = ' ';
arr['x3'] = ' ';
arr['y3'] = ' ';
arr['z3'] = ' ';
console.log(arr['z1']);
arr['z1'] = 'foo';
console.log(arr['z1']);
answered Nov 12 at 4:37
Jim B.
2,6561928
2,6561928
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
add a comment |
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54
add a comment |
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For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22
use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25
@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28
As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30
@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30