I'm using mongoose.js to do queries to mongodb, but I think my problem is not specific to mongoose.js.
Say I have only one record in the collection:
var album = new Album({
tracks: [{
title: 'track0',
language: 'en',
},{
title: 'track1',
language: 'en',
},{
title: 'track2',
language: 'es',
}]
})
I want to select all tracks with language field equal to 'en', so I tried two variants:
Album.find({'tracks.language':'en'}, {'tracks.$': 1}, function(err, albums){
and tied to to the same thing with $elemMatch projection:
Album.find({}, {tracks: {$elemMatch: {'language': 'en'}}}, function(err, albums){
in either case I've got the same result:
{tracks:[{title: 'track0', language: 'en'}]}
selected album.tracks contain only ONE track element with title 'track0' (but there should be both 'track0', 'track1'):
{tracks:[{title: 'track0', language: 'en'}, {title: 'track1', language: 'en'}]}
What am I doing wrong?
Like #JohnnyHK already said, you'll have to use the aggregation framework to accomplish that because both $ and $elemMatch only return the first match.
Here's how:
db.Album.aggregate(
// This is optional. It might make your query faster if you have
// many albums that don't have any English tracks. Take a larger
// collection and measure the difference. YMMV.
{ $match: {tracks: {$elemMatch: {'language': 'en'}} } },
// This will create an 'intermediate' document for each track
{ $unwind : "$tracks" },
// Now filter out the documents that don't contain an English track
// Note: at this point, documents' 'tracks' element is not an array
{ $match: { "tracks.language" : "en" } },
// Re-group so the output documents have the same structure, ie.
// make tracks a subdocument / array again
{ $group : { _id : "$_id", tracks : { $addToSet : "$tracks" } }}
);
You might want to try that aggregate query with only the first expression and then add expressions line by line to see how the output is changed. It's particularly important to understand how $unwind creates intermediate documents that are later re-merged using $group and $addToSet.
Results:
> db.Album.aggregate(
{ $match: {tracks: {$elemMatch: {'language': 'en'}} } },
{ $unwind : "$tracks" },
{ $match: { "tracks.language" : "en" } },
{ $group : { _id : "$_id", tracks : { $addToSet : "$tracks" } }} );
{
"result" : [
{
"_id" : ObjectId("514217b1c99766f4d210c20b"),
"tracks" : [
{
"title" : "track1",
"language" : "en"
},
{
"title" : "track0",
"language" : "en"
}
]
}
],
"ok" : 1
}
Related
I have a mongodb collection with multiple documents. Each document has an array with multiple subdocuments (or embedded documents i guess?). Each of these subdocuments is in this format:
{
"name": string,
"count": integer
}
Now I want to aggregate these subdocuments to find
The top X counts and their name.
Same as 1. but the names have to match a regex before sorting and limiting.
I have tried the following for 1. already - it does return me the top X but unordered, so I'd have to order them again which seems somewhat inefficient.
[{
$match: {
_id: id
}
}, {
$unwind: {
path: "$array"
}
}, {
$sort: {
'count': -1
}
}, {
$limit: x
}]
Since i'm rather new to mongodb this is pretty confusing for me. Happy for any help. Thanks in advance.
The sort has to include the array name in order to avoid an additional sort later on.
Given the following document to work with:
{
students: [{
count: 4,
name: "Ann"
}, {
count: 7,
name: "Brad"
}, {
count: 6,
name: "Beth"
}, {
count: 8,
name: "Catherine"
}]
}
As an example, the following aggregation query will match any name containing the letters "h" and "e". This needs to happen after the "$unwind" step in order to only keep the ones you need.
db.tests.aggregate([
{$match: {
_id: ObjectId("5c1b191b251d9663f4e3ce65")
}},
{$unwind: {
path: "$students"
}},
{$match: {
"students.name": /[he]/
}},
{$sort: {
"students.count": -1
}},
{$limit: 2}
])
This is the output given the above mentioned input:
{ "_id" : ObjectId("5c1b191b251d9663f4e3ce65"), "students" : { "count" : 8, "name" : "Catherine" } }
{ "_id" : ObjectId("5c1b191b251d9663f4e3ce65"), "students" : { "count" : 6, "name" : "Beth" } }
Both names contain the letters "h" and "e", and the output is sorted from high to low.
When setting the limit to 1, the output is limited to:
{ "_id" : ObjectId("5c1b191b251d9663f4e3ce65"), "students" : { "count" : 8, "name" : "Catherine" } }
In this case only the highest count has been kept after having matched the names.
=====================
Edit for the extra question:
Yes, the first $match can be changed to filter on specific universities.
{$match: {
university: "University X"
}},
That will give one or more matching documents (in case you have a document per year or so) and the rest of the aggregation steps would still be valid.
The following match would retrieve the students for the given university for a given academic year in case that would be needed.
{$match: {
university: "University X",
academic_year: "2018-2019"
}},
That should narrow it down to get the correct documents.
I have a user document like:
{
_id: "s0m3Id",
_skills: ["skill1", "skill2"],
}
Now I want to unwind this document by the _skills field and add a score for each skill. So my aggregate looks like:
{
'$unwind': {'path': '$_skills', 'preserveNullAndEmptyArrays': true},
},
{
'$project': {
'_skills':
'label': '$_skills',
'skill_score': 1
},
}
},
Sometimes the _skills field can be empty, however in this case I still want the user document to flow through the aggregation - hence the preserveNullAndEmptyArrays parameter. However, the problem I'm having is that it will project a skill_score (though with no label) onto documents which had empty _skills array fields. Thus, when I go to $group the documents later on, those documents now have a non-empty _skills array, containing a single object, namely {skill_score: 1}. This is not what I want - I want documents which had empty (or non-existent) _skills fields to not have any skill_score projected onto them.
So how can I conditionally project a field based on the existence of another field? Using $exists does not help, because that is intended for querying, not for boolean expressions.
Updated
This aggregation will set the value of skill_score to 0 if _skills does not exist, then use $redact to remove the subdocument having skill_score equals to 0:
db.project_if.aggregate([
{
$unwind: {
path: '$_skills',
preserveNullAndEmptyArrays: true,
}
},
{
$project: {
_skills: {
label: '$_skills',
skill_score: {
$cond: {
if: {
$eq: ['$_skills', undefined]
},
then: 0,
else: 1,
}
}
}
}
},
{
$redact: {
$cond: {
if: { $eq: [ "$skill_score", 0 ] },
then: '$$PRUNE',
else: '$$DESCEND'
}
}
}
]);
Result would be like:
[
{ "_id" : '', "_skills" : { "label" : "skill1", "skill_score" : 1 } },
{ "_id" : '', "_skills" : { "label" : "skill2", "skill_score" : 1 } },
{ "_id" : '' },
]
I have the following documents:
[{
"_id":1,
"name":"john",
"position":1
},
{"_id":2,
"name":"bob",
"position":2
},
{"_id":3,
"name":"tom",
"position":3
}]
In the UI a user can change position of items(eg moving Bob to first position, john gets position 2, tom - position 3).
Is there any way to update all positions in all documents at once?
You can not update two documents at once with a MongoDB query. You will always have to do that in two queries. You can of course set a value of a field to the same value, or increment with the same number, but you can not do two distinct updates in MongoDB with the same query.
You can use db.collection.bulkWrite() to perform multiple operations in bulk. It has been available since 3.2.
It is possible to perform operations out of order to increase performance.
From mongodb 4.2 you can do using pipeline in update using $set operator
there are many ways possible now due to many operators in aggregation pipeline though I am providing one of them
exports.updateDisplayOrder = async keyValPairArr => {
try {
let data = await ContestModel.collection.update(
{ _id: { $in: keyValPairArr.map(o => o.id) } },
[{
$set: {
displayOrder: {
$let: {
vars: { obj: { $arrayElemAt: [{ $filter: { input: keyValPairArr, as: "kvpa", cond: { $eq: ["$$kvpa.id", "$_id"] } } }, 0] } },
in:"$$obj.displayOrder"
}
}
}
}],
{ runValidators: true, multi: true }
)
return data;
} catch (error) {
throw error;
}
}
example key val pair is: [{"id":"5e7643d436963c21f14582ee","displayOrder":9}, {"id":"5e7643e736963c21f14582ef","displayOrder":4}]
Since MongoDB 4.2 update can accept aggregation pipeline as second argument, allowing modification of multiple documents based on their data.
See https://docs.mongodb.com/manual/reference/method/db.collection.update/#modify-a-field-using-the-values-of-the-other-fields-in-the-document
Excerpt from documentation:
Modify a Field Using the Values of the Other Fields in the Document
Create a members collection with the following documents:
db.members.insertMany([
{ "_id" : 1, "member" : "abc123", "status" : "A", "points" : 2, "misc1" : "note to self: confirm status", "misc2" : "Need to activate", "lastUpdate" : ISODate("2019-01-01T00:00:00Z") },
{ "_id" : 2, "member" : "xyz123", "status" : "A", "points" : 60, "misc1" : "reminder: ping me at 100pts", "misc2" : "Some random comment", "lastUpdate" : ISODate("2019-01-01T00:00:00Z") }
])
Assume that instead of separate misc1 and misc2 fields, you want to gather these into a new comments field. The following update operation uses an aggregation pipeline to:
add the new comments field and set the lastUpdate field.
remove the misc1 and misc2 fields for all documents in the collection.
db.members.update(
{ },
[
{ $set: { status: "Modified", comments: [ "$misc1", "$misc2" ], lastUpdate: "$$NOW" } },
{ $unset: [ "misc1", "misc2" ] }
],
{ multi: true }
)
Suppose after updating your position your array will looks like
const objectToUpdate = [{
"_id":1,
"name":"john",
"position":2
},
{
"_id":2,
"name":"bob",
"position":1
},
{
"_id":3,
"name":"tom",
"position":3
}].map( eachObj => {
return {
updateOne: {
filter: { _id: eachObj._id },
update: { name: eachObj.name, position: eachObj.position }
}
}
})
YourModelName.bulkWrite(objectToUpdate,
{ ordered: false }
).then((result) => {
console.log(result);
}).catch(err=>{
console.log(err.result.result.writeErrors[0].err.op.q);
})
It will update all position with different value.
Note : I have used here ordered : false for better performance.
Lets say I have mongo documents like this:
Question 1
{
answers:[
{content: 'answer1'},
{content: '2nd answer'}
]
}
Question 2
{
answers:[
{content: 'answer1'},
{content: '2nd answer'}
{content: 'The third answer'}
]
}
Is there a way to order the collection by size of answers?
After a little research I saw suggestions of adding another field, that would contain number of answers and use it as a reference but may be there is native way to do it?
I thought you might be able to use $size, but that's only to find arrays of a certain size, not ordering.
From the mongo documentation:
http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24size
You cannot use $size to find a range of sizes (for example: arrays with more than 1 element). If you need to query for a range, create an extra size field that you increment when you add elements. Indexes cannot be used for the $size portion of a query, although if other query expressions are included indexes may be used to search for matches on that portion of the query expression.
Looks like you can probably fairly easily do this with the new aggregation framework, edit: which isn't out yet.
http://www.mongodb.org/display/DOCS/Aggregation+Framework
Update Now the Aggregation Framework is out...
> db.test.aggregate([
{$unwind: "$answers"},
{$group: {_id:"$_id", answers: {$push:"$answers"}, size: {$sum:1}}},
{$sort:{size:1}}]);
{
"result" : [
{
"_id" : ObjectId("5053b4547d820880c3469365"),
"answers" : [
{
"content" : "answer1"
},
{
"content" : "2nd answer"
}
],
"size" : 2
},
{
"_id" : ObjectId("5053b46d7d820880c3469366"),
"answers" : [
{
"content" : "answer1"
},
{
"content" : "2nd answer"
},
{
"content" : "The third answer"
}
],
"size" : 3
}
],
"ok" : 1
}
I use $project for this:
db.test.aggregate([
{
$project : { answers_count: {$size: { "$ifNull": [ "$answers", [] ] } } }
},
{
$sort: {"answers_count":1}
}
])
It also allows to include documents with empty answers.
But also has a disadvantage (or sometimes advantage): you should manually add all needed fields in $project step.
you can use mongodb aggregation stage $addFields which will add extra field to store count and then followed by $sort stage.
db.test.aggregate([
{
$addFields: { answers_count: {$size: { "$ifNull": [ "$answers", [] ] } } }
},
{
$sort: {"answers_count":1}
}
])
You can use $size attribute to order by array length.
db.getCollection('test').aggregate([
{$project: { "answers": 1, "answer_count": { $size: "$answers" } }},
{$sort: {"answer_count": -1}}])
This is the documents structure:
{
'_id' : ObjectId('56be1b51a0f4c8591f37f62a'),
'name': 'Bob',
'sub_users': [{'_id' : ObjectId('56be1b51a0f4c8591f37f62a')}]
}
{
'_id' : ObjectId('56be1b51a0f4c8591f37f62b'),
'name': 'Alice',
'sub_users': [{'_id' : ObjectId('56be1b51a0f4c8591f37f62a')}]
}
The sub_users array is used basically to link accounts, in the example Alice is Bob's manager since she has him as a sub_user. Bob has his own id in the sub_users array and this is wrong (no one really is his own boss).
I want to find all the Bobs, it feels like a simple query but I can't find the way to do it, or to even to google it properly, tried this (probably knowing it wouldn't work);
db.users.aggregate([
{ $group: { _id: '_id' } },
{ $match: { sub_users: { $elemMatch: { _id: '$$ROOT._id' } } } }
])
And it didn't worked, so the question is; how to find a document whose nested documents have the same value as the root element (for a certain field)?
To get there I'm using compare expression - please see example below:
db.users.aggregate([{
$unwind : "$sub_users"
}, //have all ids on same level
{
$project : {
_id : 1,
name : 1,
sameId : {
$cmp : ["$_id", "$sub_users._id"]
},
}
}, {
$match : {
sameId : 0
}
}
])