Get specific object in array of array in MongoDB - mongodb

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.

Related

MongoDB get all embedded documents where condition is met

I did this in my mongodb:
db.teams.insert({name:"Alpha team",employees:[{name:"john"},{name:"david"}]});
db.teams.insert({name:"True team",employees:[{name:"oliver"},{name:"sam"}]});
db.teams.insert({name:"Blue team",employees:[{name:"jane"},{name:"raji"}]});
db.teams.find({"employees.name":/.*o.*/});
But what I got was:
{ "_id" : ObjectId("5ddf3ca83c182cc5354a15dd"), "name" : "Alpha team", "employees" : [ { "name" : "john" }, { "name" : "david" } ] }
{ "_id" : ObjectId("5ddf3ca93c182cc5354a15de"), "name" : "True team", "employees" : [ { "name" : "oliver" }, { "name" : "sam" } ] }
But what I really want is
[{"name":"john"},{"name":"oliver"}]
I'm having a hard time finding examples of this without using some kind of programmatic iterator/loop. Or examples I find return the parent document, which means I'd have to parse out the embedded array employees and do some kind of UNION statement?
Eg.
How to get embedded document in mongodb?
Retrieve only the queried element in an object array in MongoDB collection
Can someone point me in the right direction?
Please add projections to filter out the fields you don't need. Please refer the project link mongodb projections
Your find query should be constructed with the projection parameters like below:
db.teams.find({"employees.name":/.*o.*/}, {_id:0, "employees.name": 1});
This will return you:
[{"name":"john"},{"name":"oliver"}]
Can be solved with a simple aggregation pipeline.
db.teams.aggregate([
{$unwind : "$employees"},
{$match : {"employees.name":/.*o.*/}},
])
EDIT:
OP Wants to skip the parent fields. Modified query:
db.teams.aggregate([
{$unwind : "$employees"},
{$match : {"employees.name":/.*o.*/}},
{$project : {"name":"$employees.name",_id:0}}
])
Output:
{ "name" : "john" }
{ "name" : "oliver" }

Mongodb aggregate lookup return only one field of array

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.

MongoDB .Net driver 2.0 Builders Filter (field to array comparison)

I need to get all usernames from "followingList.username" and compare with
posts' usernames, if there any match need to add that one to an array.
Person Model
{
"_id" : ObjectId("554f20f5c90d3c7ed42303e1"),
"username" : "fatihyildizhan",
"followingList" : [
{
"_id" : ObjectId("55505b6ca515860cbcf7901d"),
"username" : "gumusluk",
"avatar" : "avatar.png"
},
{
"_id" : ObjectId("58505b6ca515860cbcf7901d"),
"username" : "yalikavak",
"avatar" : "avatar.png"
},
{
"_id" : ObjectId("58305b6ca515860cbcf7901d"),
"username" : "gumbet",
"avatar" : "avatar.png"
}
]
}
Post Model
{
"_id" : ObjectId("554f2df2a388R4b425b89833"),
"username" : "yalikavak",
"category" : "Summer",
"text" : "blue voyage with yacht"
},
{
"_id" : ObjectId("554f2df2a388P4b425b89833"),
"username" : "yalikavak",
"category" : "Winter",
"text" : "is coming ..."
},
{
"_id" : ObjectId("554f2df2a388K4b425b89833"),
"username" : "gumbet",
"category" : "Fall",
"text" : "there are many trees"
}
I try to get result code block as below but couldn't figure it out.
var filter = Builders<Post>.Filter.AnyEq("username", usernameList);
var result = collection.Find(filter).ToListAsync().Result;
Can you help me with this?
Thanks.
The logic is flipped, what you need is an $in query if I understand your use case correctly:
var filter = Builders<Post>.Filter.In("username", usernameList);
var result = collection.Find(filter).ToListAsync().Result;
In your case, username is a simple field and you want to match against a list of candidates. AnyEq is used to check that, from an embedded list of complex objects, at least one matches a criterion. That still translates to a simple query in MongoDB, but requires to 'reach into' the object which needs a more complicated syntax.

MongoDb - Query for specific subdocument

I have a set of mongodb documents with the following structure:
{
"_id" : NUUID("58fbb893-dfe9-4f08-a761-5629d889647d"),
"Identifiers" : {
"IdentificationLevel" : 2,
"Identifier" : "extranet\\test#test.com"
},
"Personal" : {
"FirstName" : "Test",
"Surname" : "Test"
},
"Tags" : {
"Entries" : {
"ContactLists" : {
"Values" : {
"0" : {
"Value" : "{292D8695-4936-4865-A413-800960626E6D}",
"DateTime" : ISODate("2015-04-30T09:14:45.549Z")
}
}
}
}
}
}
How can I make a query with the mongo shell which finds all documents with a specific "Value" (e.g.{292D8695-4936-4865-A413-800960626E6D} in the Tag.Entries.ContactLists.Values path?
The structure is unfortunately locked by Sitecore, so it is not an options to use another structure.
As your sample collection structure show Values is object, it contains only one Value. Also you must check for Value as it contains extra paranthesis. If you want to get Value from given structure try following query :
db.collection.find({
"Tags.Entries.ContactLists.Values.0.Value": "{292D8695-4936-4865-A413-800960626E6D}"
})

Error in mongodb query to get movie based on id

> db.movmodels.findOne()
{
"_id" : ObjectId("55320b0e0e9e0d9d0540593c"),
"username" : "punk",
"favMovies" : [
{
"alternate_ids" : {
"imdb" : "0137523"
},
"abridged_cast" : [
{
"characters" : [
"Tyler"
],
"id" : "162652627",
"name" : "Brad Pitt"
},
{
"characters" : [
"Narrator"
],
"id" : "162660884",
"name" : "Edward Norton"
},
{
"characters" : [
"Robert"
],
"id" : "162676383",
"name" : "Meat Loaf"
},
{
"characters" : [
"Angel Face"
],
"id" : "162653925",
"name" : "Jared Leto"
},
{
"characters" : [
"Boss"
],
"id" : "770706064",
"name" : "Zach Grenier"
}
],
"synopsis" : "",
"ratings" : {
"audience_score" : 96,
"audience_rating" : "Upright",
"critics_score" : 80,
"critics_rating" : "Certified Fresh"
},
"release_dates" : {
"dvd" : "2000-06-06",
"theater" : "1999-10-15"
},
"critics_consensus" : "",
"runtime" : 139,
"mpaa_rating" : "R",
"year" : 1999,
"title" : "Fight Club",
**"id" : "13153"**
}
],
"__v" : 0
}
This is my data in mongodb.
As I am new to mongodb I wanted to know query to get movie with a particular id.
The query that I tried is. I need to get the movie based on id so that I can remove it from my database
db.movmodels.findOne({username:"punk"},{favMovies:{id:13153}})
but this gives me error.
2015-04-18T05:41:26.221-0400 E QUERY Error: error: {
"$err" : "Can't canonicalize query: BadValue ported projection option: favMovies: { id: 13153.0 }",
"code" : 17287
}
at Error (<anonymous>)
at DBQuery.next (src/mongo/shell/query.js:259:15)
at DBCollection.findOne (src/mongo/shell/collection.js:188:22)
at (shell):1:14 at src/mongo/shell/query.js:259
There are several problems with your query:
The second parameter to find() is a projection, not part of the query. What you want is to supply one document for the query that has two properties: {"username" : "punk", favMovies : { ... } }
However, you also don't want to compare the entire sub-document favMovies, but you only want to match on one of its properties, the id, which requires to 'reach into the object' using the dot operator: {username:"punk", "favMovies.id" : 13153}.
However, that will probably not work yet, because 13153 is not the same as "13153", the latter being a string while the former is a number in JSON.
db.movmodels.findOne({username:"punk", "favMovies.id" : "13153"})
Keep in mind, however, that this will find the entire document for the user named "punk". I'm not sure what exactly your data structure should look like, but it appears you'll have to $pull the movie from the user. In general, I'd say you're embedding too much data into the user, but that's hard to tell without knowing the exact use case.
Here you go:
If you just wanted to get first user who has this fav movie:
db.movmodels.findOne({"favMovies.id": 13153});
And, if you want to know if that user has that movie as favorite.
db.movmodels.findOne({"favMovies.id": 13153, username:"punk"});
Second argument in the findOne is used to only return particular field.
You can use also $elemMatch projection operator (not to be confused with the $elemMatch query operator)
db.movmodels.find({username:"punk"},{favMovies:{$elemMatch:{id:"13153"}}});
`
If you want to find a movie that has another movie (with id 13153) in 'favMovies' array, then write the query as below:
db.movmodels.findOne({username:"punk",'favMovies.id':13153})
And if you want to find a movie with _id 55320b0e0e9e0d9d0540593cwrite the following query:
db.movmodels.findOne({username:"punk",'_id':ObjectId("55320b0e0e9e0d9d0540593c")})