MongoDB Aggregate - Query to get most recent item in group
0
Given this source data:
"_id" : ObjectId("1"), "productID" : 1, "amount" : 1, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("2"), "productID" : 2, "amount" : 2, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("3"), "productID" : 3, "amount" : 3, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("6"), "productID" : 2, "amount" : 22, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
I want to get the most recent record for each product (keyed on productID) and print out the full row for that record per product sorted on productID. So my desired output from above would be:
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
I'm new to Mongo and having problems, managed to get most of it but couldn't delete the other older records for each product.
mongodb mongodb-query aggregation-framework
edited Feb 19 '17 at 10:36
sazzad
1,89461320
asked Feb 19 '17 at 9:23
sub123sub123
133
add a comment |
0
Given this source data:
"_id" : ObjectId("1"), "productID" : 1, "amount" : 1, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("2"), "productID" : 2, "amount" : 2, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("3"), "productID" : 3, "amount" : 3, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("6"), "productID" : 2, "amount" : 22, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
I want to get the most recent record for each product (keyed on productID) and print out the full row for that record per product sorted on productID. So my desired output from above would be:
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
I'm new to Mongo and having problems, managed to get most of it but couldn't delete the other older records for each product.
mongodb mongodb-query aggregation-framework
edited Feb 19 '17 at 10:36
sazzad
1,89461320
asked Feb 19 '17 at 9:23
sub123sub123
133
add a comment |
0
0
0
Given this source data:
"_id" : ObjectId("1"), "productID" : 1, "amount" : 1, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("2"), "productID" : 2, "amount" : 2, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("3"), "productID" : 3, "amount" : 3, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("6"), "productID" : 2, "amount" : 22, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
I want to get the most recent record for each product (keyed on productID) and print out the full row for that record per product sorted on productID. So my desired output from above would be:
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
I'm new to Mongo and having problems, managed to get most of it but couldn't delete the other older records for each product.
mongodb mongodb-query aggregation-framework
edited Feb 19 '17 at 10:36
sazzad
1,89461320
asked Feb 19 '17 at 9:23
sub123sub123
133
Given this source data:
"_id" : ObjectId("1"), "productID" : 1, "amount" : 1, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("2"), "productID" : 2, "amount" : 2, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("3"), "productID" : 3, "amount" : 3, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("6"), "productID" : 2, "amount" : 22, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
I want to get the most recent record for each product (keyed on productID) and print out the full row for that record per product sorted on productID. So my desired output from above would be:
"_id" : ObjectId("5"), "productID" : 1, "amount" : 11, "date" : ISODate("2017-02-02T00:00:00Z")
"_id" : ObjectId("7"), "productID" : 2, "amount" : 222, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("8"), "productID" : 3, "amount" : 33, "date" : ISODate("2017-02-03T00:00:00Z")
"_id" : ObjectId("4"), "productID" : 4, "amount" : 4, "date" : ISODate("2017-02-01T00:00:00Z")
I'm new to Mongo and having problems, managed to get most of it but couldn't delete the other older records for each product.
mongodb mongodb-query aggregation-framework
mongodb mongodb-query aggregation-framework
edited Feb 19 '17 at 10:36
sazzad
1,89461320
asked Feb 19 '17 at 9:23
sub123sub123
133
edited Feb 19 '17 at 10:36
sazzad
1,89461320
asked Feb 19 '17 at 9:23
sub123sub123
133
edited Feb 19 '17 at 10:36
sazzad
1,89461320
edited Feb 19 '17 at 10:36
sazzad
1,89461320
edited Feb 19 '17 at 10:36
sazzad
1,89461320
1,89461320
asked Feb 19 '17 at 9:23
sub123sub123
133
asked Feb 19 '17 at 9:23
sub123sub123
133
asked Feb 19 '17 at 9:23
sub123sub123
133
133
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
You can use $last and $first operators. Note that you should sort documents before grouping:
db.so.aggregate([
$sort: productID:-1, date: -1 ,
$group:
_id : "$productID",
date: $last: "$date" ,
amount: $first: "$amount" ,
id : $first: "$_id"
,
$project: _id: "$id", productId: "$_id", date: 1, amount: 1
])
Output:
"_id" : 5,
"productId" : 1,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 11
,
"_id" : 7,
"productId" : 2,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 222
,
"_id" : 8,
"productId" : 3,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 33
,
"_id" : 4,
"productId" : 4,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 4
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You can use $last and $first operators. Note that you should sort documents before grouping:
db.so.aggregate([
$sort: productID:-1, date: -1 ,
$group:
_id : "$productID",
date: $last: "$date" ,
amount: $first: "$amount" ,
id : $first: "$_id"
,
$project: _id: "$id", productId: "$_id", date: 1, amount: 1
])
Output:
"_id" : 5,
"productId" : 1,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 11
,
"_id" : 7,
"productId" : 2,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 222
,
"_id" : 8,
"productId" : 3,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 33
,
"_id" : 4,
"productId" : 4,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 4
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
add a comment |
1
You can use $last and $first operators. Note that you should sort documents before grouping:
db.so.aggregate([
$sort: productID:-1, date: -1 ,
$group:
_id : "$productID",
date: $last: "$date" ,
amount: $first: "$amount" ,
id : $first: "$_id"
,
$project: _id: "$id", productId: "$_id", date: 1, amount: 1
])
Output:
"_id" : 5,
"productId" : 1,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 11
,
"_id" : 7,
"productId" : 2,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 222
,
"_id" : 8,
"productId" : 3,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 33
,
"_id" : 4,
"productId" : 4,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 4
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
add a comment |
1
1
1
You can use $last and $first operators. Note that you should sort documents before grouping:
db.so.aggregate([
$sort: productID:-1, date: -1 ,
$group:
_id : "$productID",
date: $last: "$date" ,
amount: $first: "$amount" ,
id : $first: "$_id"
,
$project: _id: "$id", productId: "$_id", date: 1, amount: 1
])
Output:
"_id" : 5,
"productId" : 1,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 11
,
"_id" : 7,
"productId" : 2,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 222
,
"_id" : 8,
"productId" : 3,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 33
,
"_id" : 4,
"productId" : 4,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 4
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
You can use $last and $first operators. Note that you should sort documents before grouping:
db.so.aggregate([
$sort: productID:-1, date: -1 ,
$group:
_id : "$productID",
date: $last: "$date" ,
amount: $first: "$amount" ,
id : $first: "$_id"
,
$project: _id: "$id", productId: "$_id", date: 1, amount: 1
])
Output:
"_id" : 5,
"productId" : 1,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 11
,
"_id" : 7,
"productId" : 2,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 222
,
"_id" : 8,
"productId" : 3,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 33
,
"_id" : 4,
"productId" : 4,
"date" : ISODate("2017-02-01T00:00:00.000Z"),
"amount" : 4
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
edited Feb 19 '17 at 9:42
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
answered Feb 19 '17 at 9:36
Sergey BerezovskiySergey Berezovskiy
187k23315364
187k23315364
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
add a comment |
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
Thanks that's a big help. One thing though - I am not aware of the full document schema in advance (ie this is generic)- so I know I will have a fixed date field and a id field (date, productID) but there will be many other fields that are undefined at runtime - how could I return all the fields from an object without specifying them.
– sub123
Feb 19 '17 at 21:03
add a comment |
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