Related
i have some collections for our project.
Casts collection contains movie casts
Contents collection contains movie contents
i want to run aggregate lookup for get information about movie casts with position type.
i removed collections details unnecessary fields.
Casts details:
{
"_id" : ObjectId("5a6cf47415621604942386cd"),
"fa_name" : "",
"en_name" : "Ehsan",
"fa_bio" : "",
"en_bio" : ""
}
Contents details:
{
"_id" : ObjectId("5a6b8b734f1408137f79e2cc"),
"casts" : [
{
"_id" : ObjectId("5a6cf47415621604942386cd"),
"fa_fictionName" : "",
"en_fictionName" : "Ehsan2",
"positionType" : {
"id" : 3,
"fa_name" : "",
"en_name" : "Director"
}
},
{
"_id" : ObjectId("5a6cf47415621604942386cd"),
"fa_fictionName" : "",
"en_fictionName" : "Ehsan1",
"positionType" : {
"id" : 3,
"fa_name" : "",
"en_name" : "Writers"
}
}
],
"status" : 0,
"created" : Timestamp(1516997542, 4),
"updated" : Timestamp(1516997542, 5)
}
when i run aggregate lookup with bellow query, in new generated lookup array only one casts contents If in accordance with above casts array value aggregate lookup should return two casts content with two type. in casts array value exists two type of casts, 1) writers and directors. but returned director casts content. _casts should contains two object not one object!
aggregate lookup query:
{$lookup:{from:"casts",localField:"casts._id",foreignField:"_id",as:"_casts"}}
result:
{
"_id" : ObjectId("5a6b8b734f1408137f79e2cc"),
"casts" : [
{
"_id" : ObjectId("5a6cf47415621604942386cd"),
"fa_fictionName" : "",
"en_fictionName" : "Ehsan2",
"positionType" : {
"id" : 3,
"fa_name" : "",
"en_name" : "Director"
}
},
{
"_id" : ObjectId("5a6cf47415621604942386cd"),
"fa_fictionName" : "",
"en_fictionName" : "Ehsan1",
"positionType" : {
"id" : 3,
"fa_name" : "",
"en_name" : "Writers"
}
}
],
"_casts" : [
{
"_id" : ObjectId("5a6cf47415621604942386cd"),
"fa_name" : "",
"en_name" : "Ehsan",
"fa_bio" : "",
"en_bio" : ""
}
],
"status" : 0,
"created" : Timestamp(1516997542, 4),
"updated" : Timestamp(1516997542, 5)
}
EDIT-1
finally my problem is solved. i have only one problem with this query, this query doesn't show root document fields. finally solve this problem. finally query exists in EDIT-2.
query:
db.contents.aggregate([
{"$unwind":"$casts"},
{"$lookup":{"from":"casts","localField":"casts._id","foreignField":"_id","as":"casts.info"}},
{"$unwind":"$casts.info"},
{"$group":{"_id":"$_id", "casts":{"$push":"$casts"}}},
])
EDIT-2
db.contents.aggregate([
{"$unwind":"$casts"},
{"$lookup":{"from":"casts","localField":"casts._id","foreignField":"_id","as":"casts.info"}},
{"$unwind":"$casts.info"},
{$group:{"_id":"$_id", "data":{"$first":"$$ROOT"}, "casts":{"$push":"$casts"}}},
{$replaceRoot:{"newRoot":{"$mergeObjects":["$data",{"casts":"$casts"}]}}},
{$project:{"casts":0}}
]).pretty()
This is expected behavior.
From the docs,
If your localField is an array, you may want to add an $unwind stage
to your pipeline. Otherwise, the equality condition between the
localField and foreignField is foreignField: { $in: [
localField.elem1, localField.elem2, ... ] }.
So to join each local field array element with foreign field element you have to $unwind the local array.
db.content.aggregate([
{"$unwind":"$casts"},
{"$lookup":{"from":"casts","localField":"casts._id","foreignField":"_id","as":"_casts"}}
])
Vendor Collection
Items Collection
db.items.aggregate([
{ $match:
{"item_id":{$eq:"I001"}}
},
{
$lookup:{
from:"vendor",
localField:"vendor_id",
foreignField:"vendor_id",
as:"vendor_details"
}
},
{
$unwind:"$vendor_details"
},
{
$project:{
"_id":0,
"vendor_id":0,
"vendor_details.vendor_company_description":0,
"vendor_details._id":0,
"vendor_details.country":0,
"vendor_details.city":0,
"vendor_details.website":0
}
}
]);
Output
Your Casts collection shows only 1 document. Your Contents collection, likewise, shows only 1 document.
This is 1 to 1 - not 1 to 2. Aggregate is working as designed.
The Contents document has 2 "casts." These 2 casts are sub-documents. Work with those as sub-documents, or re-design your collections. I don't like using sub-documents unless I know I will not need to use them as look-ups or join on them.
I would suggest you re-design your collection.
Your Contents collection (it makes me think of "Movies") could look like this:
_id
title
releaseDate
genre
etc.
You can create a MovieCasts collection like this:
_id
movieId (this is _id from Contents collection, above)
castId (this is _id from Casts collection, below)
Casts
_id
name
age
etc.
I have a document in Mongodb collection, where I want to remove an object, using title key.
I tried using $unset, but it only removes the title key not the object to which it belongs.
{
"_id" : ObjectId("576b63d49d20504c1360f688"),
"books" : [
{
"art_id" : ObjectId("574e68e5ac9fbac82489b689"),
"title" :"abc",
"price" : 40
},
{
"art_id" : ObjectId("575f9badada0500d192c53f4"),
"title" : "xyz",
"price" : 20
},
{
"art_id" : ObjectId("57458224d86b3d1561150f17"),
"title" : "def",,
"price" : 30
}
],
"user_id" : "575570c315e27d13167dfc0d"
}
To remove the entire object that contains the query object use db.remove() query.
For your case:
db.yourcollection.remove({"books.title": "abc"});
Please double check the format in which the element of array is referenced.
This removes the entire objects that contains the embedded query obj. To remove only a single object, provide it with another field to uniquely identify it.
If you only want to remove the object that contains the title field from the array but wants to keep the object that contains the array, then please use the $pull operator. This answer will be of help.
Example: if you want to remove object
{
"art_id" : ObjectId("574e68e5ac9fbac82489b689"),
"title" :"abc",
"price" : 40
}
just from the array but keep the parent object like
{
"_id" : ObjectId("576b63d49d20504c1360f688"),
"books" : [
{
"art_id" : ObjectId("575f9badada0500d192c53f4"),
"title" : "xyz",
"price" : 20
},
{
"art_id" : ObjectId("57458224d86b3d1561150f17"),
"title" : "def",,
"price" : 30
}
],
"user_id" : "575570c315e27d13167dfc0d"
}
use
db.mycollection.update(
{'_id': ObjectId("576b63d49d20504c1360f688")},
{ $pull: { "books" : { "title": "abc" } } },
false,
true
);
$unset won't remove the object from an array. The $unset operator deletes a particular field. doc.
Use $pull instead.
The $pull operator removes from an existing array all instances of a value or values that match a specified condition.
Try following query
db.collName.update({$pull : {books:{title:abc}}})
Refer $pull-doc
Hope this will help you.
And if... ¿Do I want to delete an object that is inside a document and not as an array?
{
"_id" : ObjectId("576b63d49d20504c1360f688"),
"books" : {
"574e68e5ac9fbac82489b689": {
"art_id" : ObjectId("574e68e5ac9fbac82489b689"),
"title" :"abc",
"price" : 40
},
"575f9badada0500d192c53f4": {
"art_id" : ObjectId("575f9badada0500d192c53f4"),
"title" :"xyz",
"price" : 20
},
"57458224d86b3d1561150f17": {
"art_id" : ObjectId("57458224d86b3d1561150f17"),
"title" : "def",
"price" : 30
}
},
"user_id" : "575570c315e27d13167dfc0d"
}
The solutions is this:
db.auctions.update(
{'_id': ObjectId("576b63d49d20504c1360f688")},
{$unset: {"books.574e68e5ac9fbac82489b689":
{_id: "574e68e5ac9fbac82489b689"}}})
Try using pull.
https://docs.mongodb.com/manual/reference/operator/update/pull/
$pull
The $pull operator removes from an existing array all instances of a value or values that match a specified condition.
The $pull operator has the form:
{ $pull: { <field1>: <value|condition>, <field2>: <value|condition>, ... } }
To specify a <field> in an embedded document or in an array, use dot notation.
Try in the mongo shell
db.yourcollection.remove({books:[{title:'title_you_want'}]})
Careful with the braces.
I need get a specific object in array of array in MongoDB.
I need get only the task object = [_id = ObjectId("543429a2cb38b1d83c3ff2c2")].
My document (projects):
{
"_id" : ObjectId("543428c2cb38b1d83c3ff2bd"),
"name" : "new project",
"author" : ObjectId("5424ac37eb0ea85d4c921f8b"),
"members" : [
ObjectId("5424ac37eb0ea85d4c921f8b")
],
"US" : [
{
"_id" : ObjectId("5434297fcb38b1d83c3ff2c0"),
"name" : "Test Story",
"author" : ObjectId("5424ac37eb0ea85d4c921f8b"),
"tasks" : [
{
"_id" : ObjectId("54342987cb38b1d83c3ff2c1"),
"name" : "teste3",
"author" : ObjectId("5424ac37eb0ea85d4c921f8b")
},
{
"_id" : ObjectId("543429a2cb38b1d83c3ff2c2"),
"name" : "jklasdfa_XXX",
"author" : ObjectId("5424ac37eb0ea85d4c921f8b")
}
]
}
]
}
Result expected:
{
"_id" : ObjectId("543429a2cb38b1d83c3ff2c2"),
"name" : "jklasdfa_XXX",
"author" : ObjectId("5424ac37eb0ea85d4c921f8b")
}
But i not getting it.
I still testing with no success:
db.projects.find({
"US.tasks._id" : ObjectId("543429a2cb38b1d83c3ff2c2")
}, { "US.tasks.$" : 1 })
I tryed with $elemMatch too, but return nothing.
db.projects.find({
"US" : {
"tasks" : {
$elemMatch : {
"_id" : ObjectId("543429a2cb38b1d83c3ff2c2")
}
}
}
})
Can i get ONLY my result expected using find()? If not, what and how use?
Thanks!
You will need an aggregation for that:
db.projects.aggregate([{$unwind:"$US"},
{$unwind:"$US.tasks"},
{$match:{"US.tasks._id":ObjectId("543429a2cb38b1d83c3ff2c2")}},
{$project:{_id:0,"task":"$US.tasks"}}])
should return
{ task : {
"_id" : ObjectId("543429a2cb38b1d83c3ff2c2"),
"name" : "jklasdfa_XXX",
"author" : ObjectId("5424ac37eb0ea85d4c921f8b")
}
Explanation:
$unwind creates a new (virtual) document for each array element
$match is the query part of your find
$project is similar as to project part in find i.e. it specifies the fields you want to get in the results
You might want to add a second $match before the $unwind if you know the document you are searching (look at performance metrics).
Edit: added a second $unwind since US is an array.
Don't know what you are doing (so realy can't tell and just sugesting) but you might want to examine if your schema (and mongodb) is ideal for your task because the document looks just like denormalized relational data probably a relational database would be better for you.
I have a collection of documents in mongodb, each of which have a "group" field that refers to a group that owns the document. The documents look like this:
{
group: <objectID>
name: <string>
contents: <string>
date: <Date>
}
I'd like to construct a query which returns the most recent N documents for each group. For example, suppose there are 5 groups, each of which have 20 documents. I want to write a query which will return the top 3 for each group, which would return 15 documents, 3 from each group. Each group gets 3, even if another group has a 4th that's more recent.
In the SQL world, I believe this type of query is done with "partition by" and a counter. Is there such a thing in mongodb, short of doing N+1 separate queries for N groups?
You cannot do this using the aggregation framework yet - you can get the $max or top date value for each group but aggregation framework does not yet have a way to accumulate top N plus there is no way to push the entire document into the result set (only individual fields).
So you have to fall back on MapReduce. Here is something that would work, but I'm sure there are many variants (all require somehow sorting an array of objects based on a specific attribute, I borrowed my solution from one of the answers in this question.
Map function - outputs group name as a key and the entire rest of the document as the value - but it outputs it as a document containing an array because we will try to accumulate an array of results per group:
map = function () {
emit(this.name, {a:[this]});
}
The reduce function will accumulate all the documents belonging to the same group into one array (via concat). Note that if you optimize reduce to keep only the top five array elements by checking date then you won't need the finalize function, and you will use less memory during running mapreduce (it will also be faster).
reduce = function (key, values) {
result={a:[]};
values.forEach( function(v) {
result.a = v.a.concat(result.a);
} );
return result;
}
Since I'm keeping all values for each key, I need a finalize function to pull out only latest five elements per key.
final = function (key, value) {
Array.prototype.sortByProp = function(p){
return this.sort(function(a,b){
return (a[p] < b[p]) ? 1 : (a[p] > b[p]) ? -1 : 0;
});
}
value.a.sortByProp('date');
return value.a.slice(0,5);
}
Using a template document similar to one you provided, you run this by calling mapReduce command:
> db.top5.mapReduce(map, reduce, {finalize:final, out:{inline:1}})
{
"results" : [
{
"_id" : "group1",
"value" : [
{
"_id" : ObjectId("516f011fbfd3e39f184cfe13"),
"name" : "group1",
"date" : ISODate("2013-04-17T20:07:59.498Z"),
"contents" : 0.23778377776034176
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe0e"),
"name" : "group1",
"date" : ISODate("2013-04-17T20:07:59.467Z"),
"contents" : 0.4434165076818317
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe09"),
"name" : "group1",
"date" : ISODate("2013-04-17T20:07:59.436Z"),
"contents" : 0.5935856597498059
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe04"),
"name" : "group1",
"date" : ISODate("2013-04-17T20:07:59.405Z"),
"contents" : 0.3912118375301361
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfdff"),
"name" : "group1",
"date" : ISODate("2013-04-17T20:07:59.372Z"),
"contents" : 0.221651989268139
}
]
},
{
"_id" : "group2",
"value" : [
{
"_id" : ObjectId("516f011fbfd3e39f184cfe14"),
"name" : "group2",
"date" : ISODate("2013-04-17T20:07:59.504Z"),
"contents" : 0.019611883210018277
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe0f"),
"name" : "group2",
"date" : ISODate("2013-04-17T20:07:59.473Z"),
"contents" : 0.5670706110540777
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe0a"),
"name" : "group2",
"date" : ISODate("2013-04-17T20:07:59.442Z"),
"contents" : 0.893193120136857
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe05"),
"name" : "group2",
"date" : ISODate("2013-04-17T20:07:59.411Z"),
"contents" : 0.9496864483226091
},
{
"_id" : ObjectId("516f011fbfd3e39f184cfe00"),
"name" : "group2",
"date" : ISODate("2013-04-17T20:07:59.378Z"),
"contents" : 0.013748752186074853
}
]
},
{
"_id" : "group3",
...
}
]
}
],
"timeMillis" : 15,
"counts" : {
"input" : 80,
"emit" : 80,
"reduce" : 5,
"output" : 5
},
"ok" : 1,
}
Each result has _id as group name and values as array of most recent five documents from the collection for that group name.
you need aggregation framework $group stage piped in a $limit stage...
you want also to $sort the records in some ways or else the limit will have undefined behaviour, the returned documents will be pseudo-random (the order used internally by mongo)
something like that:
db.collection.aggregate([{$group:...},{$sort:...},{$limit:...}])
here there is the documentation if you want to know more
I have a collection that stored information about devices like the following:
/* 1 */
{
"_id" : {
"startDate" : "2012-12-20",
"endDate" : "2012-12-30",
"dimensions" : ["manufacturer", "model"],
"metrics" : ["deviceCount"]
},
"data" : {
"results" : "1"
}
}
/* 2 */
{
"_id" : {
"startDate" : "2012-12-20",
"endDate" : "2012-12-30",
"dimensions" : ["manufacturer", "model"],
"metrics" : ["deviceCount", "noOfUsers"]
},
"data" : {
"results" : "2"
}
}
/* 3 */
{
"_id" : {
"dimensions" : ["manufacturer", "model"],
"metrics" : ["deviceCount", "noOfUsers"]
},
"data" : {
"results" : "3"
}
}
And I am trying to query the documents using the _id field which will be unique. The problem I am having is that when I query for all the different attributes as in:
db.collection.find({$and: [{"_id.dimensions":{ $all: ["manufacturer","model"], $size: 2}}, {"_id.metrics": { $all:["noOfUsers","deviceCount"], $size: 2}}]});
This matches 2 and 3 documents (I don't care about the order of the attributes values), but I would like to only get 3 back. How can I say that there should not be any other attributes to _id than those that I specify in the search query?
Please advise. Thanks.
Unfortunately, I think the closest you can get to narrowing your query results to just unordered _id.dimensions and unordered _id.metrics requires you to know the other possible fields in the _id subdocument field, eg. startDate and endDate.
db.collection.find({$and: [
{"_id.dimensions":{ $all: ["manufacturer","model"], $size: 2}},
{"_id.metrics": { $all:["noOfUsers","deviceCount"], $size: 2}},
{"_id.startDate":{$exists:false}},
{"_id.endDate":{$exists:false}}
]});
If you don't know the set of possible fields in _id, then the other possible solution would be to specify the exact _id that you want, eg.
db.collection.find({"_id" : {
"dimensions" : ["manufacturer", "model"],
"metrics" : ["deviceCount", "noOfUsers"]
}})
but this means that the order of _id.dimensions and _id.metrics is significant. This last query does a document match on exact BSON representation of _id.