Can I somehow add custom field with static (not computed) value?
I want to prepare objects before send and I need to remove some fields with internal information and add field with some entity ID.
For example I have collection "test" with objects like this
{_id: ObjectId(...), data: {...}}
And I need to convert it to
{data: {...}, entity_id: 54}
So how can I add entity_id: 54 without looping over result in my code?
db.test.aggregate({ $project: {_id: 0, data: 1, entity_id: ? } })
Thanks
Note that $literal was implemented in Mongo 2.6.
So now you can simply write:
db.test.aggregate(
{$project: {_id: 0, data: 1, entity_id: {$literal: 54}}})
See $literal.
edit as of 2.6 the $literal expression exists so you don't have to use the workaround now.
Original answer: I know this may sound really dumb, but you can use a "no-op" expression to "compute" what you need.
Example:
db.test.aggregate( { $project : {_id:0, data:1, entity_id: {$add: [54]} } } )
There was a proposed $literal operator for exactly this use case but it hasn't been implemented yet, you can vote for it here.
I'm trying to look up all records that match a certain condition, in this case _id being certain values, and then return only the top 2 results, sorted by the name field.
This is what I have
db.getCollection('col1').aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$sort: {fk: 1, name: -1}},
{$group: {_id: "$fk", items: {$push: "$$ROOT"} }},
{$project: {items: {$slice: ["$items", 2]} }}
])
and it works, BUT, it's not guaranteed. According to this Mongo thread $group does not guarantee document order.
This would also mean that all of the suggested solutions here and elsewhere, which recommend using $unwind, followed by $sort, and then $group, would also not work, for the same reason.
What is the best way to accomplish this with Mongo (any version)? I've seen suggestions that this could be accomplished in the $project phase, but I'm not quite sure how.
You are correct in saying that the result of $group is never sorted.
$group does not order its output documents.
Hence doing a;
{$sort: {fk: 1}}
then grouping with
{$group: {_id: "$fk", ... }},
will be a wasted effort.
But there is a silver lining with sorting before $group stage with name: -1. Since you are using $push (not an $addToSet), inserted objects will retain the order they've had in the newly created items array in the $group result. You can see this behaviour here (copy of your pipeline)
The items array will always have;
"items": [
{
..
"name": "Michael"
},
{
..
"name": "George"
}
]
in same order, therefore your nested array sort is a non-issue! Though I am unable to find an exact quote in documentation to confirm this behaviour, you can check;
this,
or this where it is confirmed.
Also, accumulator operator list for $group, where $addToSet has "Order of the array elements is undefined." in its description, whereas the similar operator $push does not, which might be an indirect evidence? :)
Just a simple modification of your pipeline where you move the fk: 1 sort from pre-$group stage to post-$group stage;
db.getCollection('col1').aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$sort: {name: -1}},
{$group: {_id: "$fk", items: {$push: "$$ROOT"} }},
{$sort: {_id: 1}},
{$project: {items: {$slice: ["$items", 2]} }}
])
should be sufficient to have the main result array order fixed as well. Check it on mongoplayground
$group doesn't guarantee the document order but it would keep the grouped documents in the sorted order for each bucket. So in your case even though the documents after $group stage are not sorted by fk but each group (items) would be sorted by name descending. If you would like to keep the documents sorted by fk you could just add the {$sort:{fk:1}} after $group stage
You could also sort by order of values passed in your match query should you need by adding a extra field for each document. Something like
db.getCollection('col1').aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$addField:{ifk:{$indexOfArray:[[1, 2],"$fk"]}}},
{$sort: {ifk: 1, name: -1}},
{$group: {_id: "$ifk", items: {$push: "$$ROOT"}}},
{$sort: {_id : 1}},
{$project: {items: {$slice: ["$items", 2]}}}
])
Update to allow array sort without group operator : I've found the jira which is going to allow sort array.
You could try below $project stage to sort the array.There maybe various way to do it. This should sort names descending.Working but a slower solution.
{"$project":{"items":{"$reduce":{
"input":"$items",
"initialValue":[],
"in":{"$let":{
"vars":{"othis":"$$this","ovalue":"$$value"},
"in":{"$let":{
"vars":{
//return index as 0 when comparing the first value with initial value (empty) or else return the index of value from the accumlator array which is closest and less than the current value.
"index":{"$cond":{
"if":{"$eq":["$$ovalue",[]]},
"then":0,
"else":{"$reduce":{
"input":"$$ovalue",
"initialValue":0,
"in":{"$cond":{
"if":{"$lt":["$$othis.name","$$this.name"]},
"then":{"$add":["$$value",1]},
"else":"$$value"}}}}
}}
},
//insert the current value at the found index
"in":{"$concatArrays":[
{"$slice":["$$ovalue","$$index"]},
["$$othis"],
{"$slice":["$$ovalue",{"$subtract":["$$index",{"$size":"$$ovalue"}]}]}]}
}}}}
}}}}
Simple example with demonstration how each iteration works
db.b.insert({"items":[2,5,4,7,6,3]});
othis ovalue index concat arrays (parts with counts) return value
2 [] 0 [],0 [2] [],0 [2]
5 [2] 0 [],0 [5] [2],-1 [5,2]
4 [5,2] 1 [5],1 [4] [2],-1 [5,4,2]
7 [5,4,2] 0 [],0 [7] [5,4,2],-3 [7,5,4,2]
6 [7,5,4,2] 1 [7],1 [6] [5,4,2],-3 [7,6,5,4,2]
3 [7,6,5,4,2] 4 [7,6,5,4],4 [3] [2],-1 [7,6,5,4,3,2]
Reference - Sorting Array with JavaScript reduce function
There is a bit of a red herring in the question as $group does guarantee that it will be processing incoming documents in order (and that's why you have to sort of them before $group to get an ordered arrays) but there is an issue with the way you propose doing it, since pushing all the documents into a single grouping is (a) inefficient and (b) could potentially exceed maximum document size.
Since you only want top two, for each of the unique fk values, the most efficient way to accomplish it is via a "subquery" using $lookup like this:
db.coll.aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$group:{_id:"$fk"}},
{$sort: {_id: 1}},
{$lookup:{
from:"coll",
as:"items",
let:{fk:"$_id"},
pipeline:[
{$match:{$expr:{$eq:["$fk","$$fk"]}}},
{$sort:{name:-1}},
{$limit:2},
{$project:{_id:0, fk:1, name:1}}
]
}}
])
Assuming you have an index on {fk:1, name:-1} as you must to get efficient sort in your proposed code, the first two stages here will use that index via DISTINCT_SCAN plan which is very efficient, and for each of them, $lookup will use that same index to filter by single value of fk and return results already sorted and limited to first two. This will be the most efficient way to do this at least until https://jira.mongodb.org/browse/SERVER-9377 is implemented by the server.
Given documents like
{
...
name:'whatever',
games: [122, 199, 201, 222]
}
db.col.aggregate({$match:{}},
{$sort:{'games.0': -1}})
doesn't sort ... no errors ... it just doesn't sort on the first array element of the games array.
Although a query with the same syntac .. works fine
col.find({}).sort({'games.0':-1})
if I change the collection so games is an array of objects like
[ {game1:198}, {game2:201} ...]
then the aggregation works using
{$sort:{'games.game1': -1}})
what am I missing to get this to work with an array of numbers?
Try unwinding the array first by applying the $unwind operator on the array, then use $sort on the deconstructed array and finally use $group to get the original documents structure:
db.coll.aggregate([
{"$unwind": "$games"},
{"$sort": {"games": 1}},
{
"$group": {
"_id": "$_id",
"name": {"$first": "$name"},
"games": {"$push": "$games"}
}
}
])
Try this:
db.coll.aggregate([
{"$unwind": "$games"},
{"$sort": {"games": -1}}
]}
I hope this will work for you as you expected.
In mongo 3.4 find sort i.e. db.col.find({}).sort({'games.0':-1}) works as expected whereas aggregation sort doesn't.
In mongo 3.6 both find and aggregation sort works as expected.
Jira issue for that: https://jira.mongodb.org/browse/SERVER-19402
I would recommend you to update your mongo version and your aggregation query will work fine.
I have a MongoDB Document like as follows
{
"_id":1,
"name":"XYZ"
ExamScores:[
{ExamName:"Maths", UnitTest:1, Score:100},
{ExamName:"Maths", UnitTest:2, Score:80},
{ExamName:"Science", UnitTest:1, Score:90}
]
}
I Need to retrieve this document so that it has to show only Maths Array. Like as follows
{
"_id":1,
"name":"XYZ"
ExamScores:[
{ExamName:"Maths", UnitTest:1, Score:100},
{ExamName:"Maths", UnitTest:2, Score:80},
]
}
How Can I Do That ?
As #karin states there is no, normal, in query method of doing this.
In version 2.2 you can use $elemMatch to project the first matching result from ExamScores but you cannot get multiple.
That being said, the aggregation framework can do this:
db.col.aggregate([
{$unwind: '$ExamScores'},
{$match: {'ExamScores.ExamName':"Maths"}},
{$group: {_id: '$_id', name: '$name', ExamScores: {$push: '$ExamScores'}}}
])
Something like that anyway.
This has been asked before MongoDB query to limit values based on condition, the only answer there says it is not possible, but that there is a request to implement that.
Can I somehow add custom field with static (not computed) value?
I want to prepare objects before send and I need to remove some fields with internal information and add field with some entity ID.
For example I have collection "test" with objects like this
{_id: ObjectId(...), data: {...}}
And I need to convert it to
{data: {...}, entity_id: 54}
So how can I add entity_id: 54 without looping over result in my code?
db.test.aggregate({ $project: {_id: 0, data: 1, entity_id: ? } })
Thanks
Note that $literal was implemented in Mongo 2.6.
So now you can simply write:
db.test.aggregate(
{$project: {_id: 0, data: 1, entity_id: {$literal: 54}}})
See $literal.
edit as of 2.6 the $literal expression exists so you don't have to use the workaround now.
Original answer: I know this may sound really dumb, but you can use a "no-op" expression to "compute" what you need.
Example:
db.test.aggregate( { $project : {_id:0, data:1, entity_id: {$add: [54]} } } )
There was a proposed $literal operator for exactly this use case but it hasn't been implemented yet, you can vote for it here.