Here's my mongodb document:
{ "FRIENDS" : [
{ "LOC" : "13.039064033333332,77.5547927", "NAME" : "A1", "PHONE_NUMBER" : "817361155" } ],
"MYID" : "349512956",
"MYLOCATION" : "13.0389541,77.5549799",
"OCCASION" : "cafe",
"_id" : ObjectId("503d87150f43159357000001"),
"average" : "13.0389541,77.5549799" }
Here's my query
> db.test_collection.find({"MYID":"349512956"},{"FRIENDS.PHONE_NUMBER":"817361155"});
and the output
{ "FRIENDS" : [ { "PHONE_NUMBER" : "817361155" } ], "_id" : ObjectId("503d87150f43159357000001") }
However, I want the entire friend document included in the result. LOC, NAME, PHONE_NUMBER. How do I get these ? Any help will be greatly appreciated. I'm searching through the monogo docs but am unable to find my answer.
Try this
// i.e., select * from things where MYID=somethig and PHONE_NUMBER="foo"
db.test_collection.find({"MYID":"349512956", "FRIENDS.PHONE_NUMBER":"817361155"});
To be more specific, the second argument selects the fields that you want outputted. See the actual implementation below:
> db.test_collection.find
function (query, fields, limit, skip, batchSize, options) {
return new DBQuery(this._mongo, this._db, this, this._fullName,
this._massageObject(query), fields, limit, skip, batchSize, options ||
this.getQueryOptions());
}
Related
I am new to MongoTemplate. I want to define a org.springframework.data.mongodb.core.query.Query and org.springframework.data.mongodb.core.query.Criteria to fetch the data from a collection which has a complex document structure.
This is the document
{
"_id" : {
"SId" : "ANBS",
"AssetId" : "ANBS_BS21",
"ST" : NumberLong(1479114000) //StartDate
},
"ET" : NumberLong(1479117599) //EndDate,
"TS" : [
NumberLong(1479114000),
NumberLong(1479114600),
NumberLong(1479115200),
NumberLong(1479115800),
NumberLong(1479116400),
NumberLong(1479117000)
],
"Tags" : {
"ActivePower" : {
"Avg" : [
16427.575,
16991.01,
16708.2016666667,
16488.335,
17230.1933333333,
15996.9783333333
]
},
"WindSpeed" : {
"Avg" : [
64.4266666666667,
60.8583333333333,
65.275,
62.8766666666667,
63.2166666666667,
63.14
]
}
}
I want to fetch all the documents for a given TimeRange(StartTime and EndTime), WHERE AssetId=? AND startTime >= ST AND endTime <= ET.
Can anybody help me with this problem? I dont want the mongo query, but I want the MongoTemplate based query and criteria. Thanks in advance.
I got a solution, it was very simple. Posting this answer so that it can help someone else.
Query query = new Query(Criteria
.where("_id.SId").is(siteId)
.and("_id.AssetId").is(assetId)
.and("_id.ST").gte(startTimeRange)
.and("ET").lte(endTimeRange));
return mongoTemplate.find(query, MyCollection.class, "mycollection");
> 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")})
how can I return a specific value for a specific document in MongoDB? For example, I have a schema that looks like:
{
"_id" : "XfCZSje7GjynvMZu7",
"createdAt" : ISODate("2015-03-23T14:52:44.084Z"),
"services" : {
"password" : {
"bcrypt" : "$2a$10$tcb01VbDMVhH03mbRdKYL.79FPj/fFMP62BDpcvpoTfF3LPgjHJoq"
},
"resume" : {
"loginTokens" : [ ]
}
},
"emails" : {
"address" : "abc123#gmu.edu",
"verified" : true
},
"profile" : {
"companyName" : "comp1",
"flagged" : true,
"phoneNum" : "7778883333"
}}
I want to return and store the value for profile.flagged specifically for the document with _id : XfCZSje7GjynvMZu7. So far I have tried:
db.users.find({_id:'myfi3E4YTf9z6tdgS'},{admin:1})
and
db.users.find({_id: 'myfi3E4YTf9z6tdgS'}, {profile:admin});
I want the query to return true or false depending on the assigned value.
Can someone help? Thanks!
MongoDB queries always return document objects, not single values. So one way to do this is with shell code like:
var flagged =
db.users.findOne({_id: 'myfi3E4YTf9z6tdgS'}, {'profile.flagged': 1}).profile.flagged;
Note the use of findOne instead of find so that you're working with just a single doc instead of the cursor that you get with find.
The correct answer here is the method .distinct() (link here)
Use it like this:
db.users.find({_id:'myfi3E4YTf9z6tdgS'},{admin:1}).distinct('admin')
The result will be: 1 or 0
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 $unwind since US is an array.
Don't know what you are doing (so realy can't tell and just sugesting) but you might want to examine if your schema (and mongodb) is ideal for your task because the document looks just like denormalized relational data probably a relational database would be better for you.
My MongoDB collection is made up of 2 main collections :
1) Maps
{
"_id" : ObjectId("542489232436657966204394"),
"fileName" : "importFile1.json",
"territories" : [
{
"$ref" : "territories",
"$id" : ObjectId("5424892224366579662042e9")
},
{
"$ref" : "territories",
"$id" : ObjectId("5424892224366579662042ea")
}
]
},
{
"_id" : ObjectId("542489262436657966204398"),
"fileName" : "importFile2.json",
"territories" : [
{
"$ref" : "territories",
"$id" : ObjectId("542489232436657966204395")
}
],
"uploadDate" : ISODate("2012-08-22T09:06:40.000Z")
}
2) Territories, which are referenced in "Map" objects :
{
"_id" : ObjectId("5424892224366579662042e9"),
"name" : "Afghanistan",
"area" : 653958
},
{
"_id" : ObjectId("5424892224366579662042ea"),
"name" : "Angola",
"area" : 1252651
},
{
"_id" : ObjectId("542489232436657966204395"),
"name" : "Unknown",
"area" : 0
}
My objective is to list every map with their cumulative area and number of territories. I am trying the following query :
db.maps.aggregate(
{'$unwind':'$territories'},
{'$group':{
'_id':'$fileName',
'numberOf': {'$sum': '$territories.name'},
'locatedArea':{'$sum':'$territories.area'}
}
})
However the results show 0 for each of these values :
{
"result" : [
{
"_id" : "importFile2.json",
"numberOf" : 0,
"locatedArea" : 0
},
{
"_id" : "importFile1.json",
"numberOf" : 0,
"locatedArea" : 0
}
],
"ok" : 1
}
I probably did something wrong when trying to access to the member variables of Territory (name and area), but I couldn't find an example of such a case in the Mongo doc. area is stored as an integer, and name as a string.
I probably did something wrong when trying to access to the member variables of Territory (name and area), but I couldn't find an example
of such a case in the Mongo doc. area is stored as an integer, and
name as a string.
Yes indeed, the field "territories" has an array of database references and not the actual documents. DBRefs are objects that contain information with which we can locate the actual documents.
In the above example, you can clearly see this, fire the below mongo query:
db.maps.find({"_id":ObjectId("542489232436657966204394")}).forEach(function(do
c){print(doc.territories[0]);})
it will print the DBRef object rather than the document itself:
o/p: DBRef("territories", ObjectId("5424892224366579662042e9"))
so, '$sum': '$territories.name','$sum': '$territories.area' would show you '0' since there are no fields such as name or area.
So you need to resolve this reference to a document before doing something like $territories.name
To achieve what you want, you can make use of the map() function, since aggregation nor Map-reduce support sub queries, and you already have a self-contained map document, with references to its territories.
Steps to achieve:
a) get each map
b) resolve the `DBRef`.
c) calculate the total area, and the number of territories.
d) make and return the desired structure.
Mongo shell script:
db.maps.find().map(function(doc) {
var territory_refs = doc.territories.map(function(terr_ref) {
refName = terr_ref.$ref;
return terr_ref.$id;
});
var areaSum = 0;
db.refName.find({
"_id" : {
$in : territory_refs
}
}).forEach(function(i) {
areaSum += i.area;
});
return {
"id" : doc.fileName,
"noOfTerritories" : territory_refs.length,
"areaSum" : areaSum
};
})
o/p:
[
{
"id" : "importFile1.json",
"noOfTerritories" : 2,
"areaSum" : 1906609
},
{
"id" : "importFile2.json",
"noOfTerritories" : 1,
"areaSum" : 0
}
]
Map-Reduce functions should not be and cannot be used to resolve DBRefs in the server side.
See what the documentation has to say:
The map function should not access the database for any reason.
The map function should be pure, or have no impact outside of the
function (i.e. side effects.)
The reduce function should not access the database, even to perform
read operations. The reduce function should not affect the outside
system.
Moreover, a reduce function even if used(which can never work anyway) will never be called for your problem, since a group w.r.t "fileName" or "ObjectId" would always have only one document, in your dataset.
MongoDB will not call the reduce function for a key that has only a
single value