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MongoDB get full doc after match, group, and sort

MongoDb/PHP, can't get grouping to work right with aggregate frameworkMongoDB - sort by subdocument matchTrouble using $group twice in mongodb aggregation queryGetting a complete object via $max in mongodb aggregationMongoDB aggregation pipeline $match orderMongoDB Aggregation Count of Group By Inner ArrayMongodb: Multi _id aggregate group not using indexWhy sort is affecting match to use indexes on MongoDB 3.0?MongoDB - Grouping by a substringMongoDB - Query - Top Document per Group

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

1

Order:

 order_id: 1,
 order_time: ISODate(...),
 customer_id: 456,
 products: [
 
 product_id: 1,
 product_name: "Pencil"
 ,
 
 product_id: 2,
 product_name: "Scissors"
 ,
 
 product_id: 3,
 product_name: "Tape"
 
 ]

I have a collection with a whole bunch of documents like the above. I would like to query for the latest order for each customer who ordered Scissors.

That is, where there exists a "products.product_name" which equals "Scissors", group by customer_id, give me the full document where the "order_time" is the "max" for that group.

To find the documents, I could do like find( 'products.product_name' : "Scissors" ) but then I get all of the order with Scissors, I only want the most recent.

So, I am looking at aggregation... Mongo's "$group" aggregation stage seems to require that you do some kind of actual aggregation inside like sum or max or whatever. I am guessing there's some combination of $match, $group, and $sort to use here but I can't seem to quite get it working.

Something close:

db.storcap.aggregate(
[

 $match: 'products.product_name' : "Scissors" 
,

 $sort: created_at:-1 
,

 $group: 
 _id: "$customer_id", 
 
]
)

But this doesn't return the full doc and I am not sure that it's doing the sorting and grouping right.

mongodb aggregation-framework

share|improve this question

asked Mar 22 at 21:03

MichaelBMichaelB

553720

add a comment | 

1

Order:

 order_id: 1,
 order_time: ISODate(...),
 customer_id: 456,
 products: [
 
 product_id: 1,
 product_name: "Pencil"
 ,
 
 product_id: 2,
 product_name: "Scissors"
 ,
 
 product_id: 3,
 product_name: "Tape"
 
 ]

I have a collection with a whole bunch of documents like the above. I would like to query for the latest order for each customer who ordered Scissors.

That is, where there exists a "products.product_name" which equals "Scissors", group by customer_id, give me the full document where the "order_time" is the "max" for that group.

To find the documents, I could do like find( 'products.product_name' : "Scissors" ) but then I get all of the order with Scissors, I only want the most recent.

So, I am looking at aggregation... Mongo's "$group" aggregation stage seems to require that you do some kind of actual aggregation inside like sum or max or whatever. I am guessing there's some combination of $match, $group, and $sort to use here but I can't seem to quite get it working.

Something close:

db.storcap.aggregate(
[

 $match: 'products.product_name' : "Scissors" 
,

 $sort: created_at:-1 
,

 $group: 
 _id: "$customer_id", 
 
]
)

But this doesn't return the full doc and I am not sure that it's doing the sorting and grouping right.

mongodb aggregation-framework

share|improve this question

asked Mar 22 at 21:03

MichaelBMichaelB

553720

add a comment | 

Order:

 order_id: 1,
 order_time: ISODate(...),
 customer_id: 456,
 products: [
 
 product_id: 1,
 product_name: "Pencil"
 ,
 
 product_id: 2,
 product_name: "Scissors"
 ,
 
 product_id: 3,
 product_name: "Tape"
 
 ]

I have a collection with a whole bunch of documents like the above. I would like to query for the latest order for each customer who ordered Scissors.

That is, where there exists a "products.product_name" which equals "Scissors", group by customer_id, give me the full document where the "order_time" is the "max" for that group.

To find the documents, I could do like find( 'products.product_name' : "Scissors" ) but then I get all of the order with Scissors, I only want the most recent.

So, I am looking at aggregation... Mongo's "$group" aggregation stage seems to require that you do some kind of actual aggregation inside like sum or max or whatever. I am guessing there's some combination of $match, $group, and $sort to use here but I can't seem to quite get it working.

Something close:

db.storcap.aggregate(
[

 $match: 'products.product_name' : "Scissors" 
,

 $sort: created_at:-1 
,

 $group: 
 _id: "$customer_id", 
 
]
)

But this doesn't return the full doc and I am not sure that it's doing the sorting and grouping right.

mongodb aggregation-framework

share|improve this question

asked Mar 22 at 21:03

MichaelBMichaelB

553720

 order_id: 1,
 order_time: ISODate(...),
 customer_id: 456,
 products: [
 
 product_id: 1,
 product_name: "Pencil"
 ,
 
 product_id: 2,
 product_name: "Scissors"
 ,
 
 product_id: 3,
 product_name: "Tape"
 
 ]

db.storcap.aggregate(
[

 $match: 'products.product_name' : "Scissors" 
,

 $sort: created_at:-1 
,

 $group: 
 _id: "$customer_id", 
 
]
)

share|improve this question

asked Mar 22 at 21:03

MichaelBMichaelB

553720

share|improve this question

asked Mar 22 at 21:03

MichaelBMichaelB

553720

asked Mar 22 at 21:03

MichaelBMichaelB

553720

asked Mar 22 at 21:03

MichaelBMichaelB

553720

1 Answer
1

active

oldest

votes

1

You can use $first operator to get most recent order (are ordered desc) and special variable $$ROOT to get whole object in a final result:

db.storcap.aggregate([
 
 $match: 'products.product_name' : "Scissors" 
 ,
 
 $sort: created_at:-1 
 ,
 
 $group: 
 _id: "$customer_id",
 lastOrder: $first: "$$ROOT" 
 
 
])

share|improve this answer

answered Mar 22 at 21:07

micklmickl

16.8k61841

add a comment | 

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1

You can use $first operator to get most recent order (are ordered desc) and special variable $$ROOT to get whole object in a final result:

db.storcap.aggregate([
 
 $match: 'products.product_name' : "Scissors" 
 ,
 
 $sort: created_at:-1 
 ,
 
 $group: 
 _id: "$customer_id",
 lastOrder: $first: "$$ROOT" 
 
 
])

share|improve this answer

answered Mar 22 at 21:07

micklmickl

16.8k61841

add a comment | 

1

You can use $first operator to get most recent order (are ordered desc) and special variable $$ROOT to get whole object in a final result:

db.storcap.aggregate([
 
 $match: 'products.product_name' : "Scissors" 
 ,
 
 $sort: created_at:-1 
 ,
 
 $group: 
 _id: "$customer_id",
 lastOrder: $first: "$$ROOT" 
 
 
])

share|improve this answer

answered Mar 22 at 21:07

micklmickl

16.8k61841

add a comment | 

You can use $first operator to get most recent order (are ordered desc) and special variable $$ROOT to get whole object in a final result:

db.storcap.aggregate([
 
 $match: 'products.product_name' : "Scissors" 
 ,
 
 $sort: created_at:-1 
 ,
 
 $group: 
 _id: "$customer_id",
 lastOrder: $first: "$$ROOT" 
 
 
])

share|improve this answer

answered Mar 22 at 21:07

micklmickl

16.8k61841

db.storcap.aggregate([
 
 $match: 'products.product_name' : "Scissors" 
 ,
 
 $sort: created_at:-1 
 ,
 
 $group: 
 _id: "$customer_id",
 lastOrder: $first: "$$ROOT" 
 
 
])

share|improve this answer

answered Mar 22 at 21:07

micklmickl

16.8k61841

answered Mar 22 at 21:07

micklmickl

16.8k61841

answered Mar 22 at 21:07

micklmickl

16.8k61841