Using embedded documents in spring data aggregation - mongodb

I have a MongoDB document like this example doc:
{
"_id" : "A",
"articleNumber" : "0123456",
"shopDependentProperties" :
{
"shop" : "DE",
"foo" : "foo",
"bar" : "bar"
}
}
and want to pull out the properties of shopDependentProperties, to have the following result
{
"_id" : "A",
"articleNumber" : "0123456",
"foo" : "foo",
"bar" : "bar"
}
In MongoDB Shell i can solve it this way:
db.test.aggregate(
[
{
$project:
{
_id : "$_id",
articleNumber : "$articleNumber",
foo:"$shopDependentProperties.foo",
bar:"$shopDependentProperties.bar"
}
}
]
)
But: In Spring Data MongoDB i can't extract the embedded document contents.
I tried many combinations, nothing worked. For example:
ProjectionOperation projection = Aggregation.project("_id");
projection.andExpression("shopDependentProperties.foo").as("foo");
projection.andExpression("shopDependentProperties.bar").as("bar");
System.out.println(projection.toDBObject(Aggregation.DEFAULT_CONTEXT));
will ignore the shopDependentProperties.shop stuff and just print out
{ "$project" : { "_id" : 1}}
Any suggestions?
Thx

Haven't tested this, but as of
http://docs.mongodb.org/manual/reference/operator/aggregation/project/
you specify included / excluded fields like this:
db.test.aggregate(
[
{
$project:
{
_id : "$_id",
articleNumber : 1,
"shopDependentProperties.foo": 1,
"shopDependentProperties.bar": 1
}
}
]
)
Further down they explain, how to include embedded documents in the projection result.

I know how to do it in MongoDB, the problem was about doing the same thing in Spring Data.
But it works the same way, why didn't I try that before?
Solution:
ProjectionOperation projection = Aggregation.project(
"brandName",
"$shopDependentProperties.foo",
"$shopDependentProperties.bar" );

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: How to get the object names in collection?

and think you in advance for the help. I have recently started using mongoDB for some personal project and I'm interested in finding a better way to query my data.
My question is: I have the following collection:
{
"_id" : ObjectId("5dbd77f7a204d21119cfc758"),
"Toyota" : {
"Founder" : "Kiichiro Toyoda",
"Founded" : "28 August 1937",
"Subsidiaries" : [
"Lexus",
"Daihatsu",
"Subaru",
"Hino"
]
}
}
{
"_id" : ObjectId("5dbd78d3a204d21119cfc759"),
"Volkswagen" : {
"Founder" : "German Labour Front",
"Founded" : "28 May 1937",
"Subsidiaries" : [
"Audi",
"Volkswagen",
"Skoda",
"SEAT"
]
}
}
I want to get the object name for example here I want to return
[Toyota, Volkswagen]
I have use this method
var names = {}
db.cars.find().forEach(function(doc){Object.keys(doc).forEach(function(key){names[key]=1})});
names;
which gave me the following result:
{ "_id" : 1, "Toyota" : 1, "Volkswagen" : 1 }
however, is there a better way to get the same result and also to just return the names of the objects. Thank you.
I would suggest you to change the schema design to be something like:
{
_id: ...,
company: {
name: 'Volkswagen',
founder: ...,
subsidiaries: ...,
...<other fields>...
}
You can then use the aggregation framework to achieve a similar result:
> db.test.find()
{ "_id" : 0, "company" : { "name" : "Volkswagen", "founder" : "German Labour Front" } }
{ "_id" : 1, "company" : { "name" : "Toyota", "founder" : "Kiichiro Toyoda" } }
> db.test.aggregate([ {$group: {_id: null, companies: {$push: '$company.name'}}} ])
{ "_id" : null, "companies" : [ "Volkswagen", "Toyota" ] }
For more details, see:
Aggregation framework
$group
Accumulator operators
As a bonus, you can create an index on the company.name field, whereas you cannot create an index on varying field names like in your example.

mongoDB distict problems

It's one of my data as JSON format:
{
"_id" : ObjectId("5bfdb412a80939b6ed682090"),
"accounts" : [
{
"_id" : ObjectId("5bf106eee639bd0df4bd8e05"),
"accountType" : "DDA",
"productName" : "DDA1"
},
{
"_id" : ObjectId("5bf106eee639bd0df4bd8df8"),
"accountType" : "VSA",
"productName" : "VSA1"
},
{
"_id" : ObjectId("5bf106eee639bd0df4bd8df9"),
"accountType" : "VSA",
"productName" : "VSA2"
}
]
}
I want to make a query to get all productName(no duplicate) of accountType = VSA.
I write a mongo query:
db.Collection.distinct("accounts.productName", {"accounts.accountType": "VSA" })
I expect: ['VSA1', 'VSA2']
I get: ['DDA','VSA1', 'VSA2']
Anybody knows why the query doesn't work in distinct?
Second parameter of distinct method represents:
A query that specifies the documents from which to retrieve the distinct values.
But the thing is that you showed only one document with nested array of elements so whole document will be returned for your condition "accounts.accountType": "VSA".
To fix that you have to use Aggregation Framework and $unwind nested array before you apply the filtering and then you can use $group with $addToSet to get unique values. Try:
db.col.aggregate([
{
$unwind: "$accounts"
},
{
$match: {
"accounts.accountType": "VSA"
}
},
{
$group: {
_id: null,
uniqueProductNames: { $addToSet: "$accounts.productName" }
}
}
])
which prints:
{ "_id" : null, "uniqueProductNames" : [ "VSA2", "VSA1" ] }

mongoDB: Querying for documents that may have some specifics options

I'm quite new to mongodb and there is one thing I can't solve right now:
Let's pretend, you have the following document structure:
{
"_id": ObjectId("some object id"),
name: "valueName",
options: [
{idOption: "optionId", name: "optionName"},
{idOption: "optionId", name: "optionName"}
]
}
And each document can have multiples options that are already classified.
I'm trying to get all the documents in the collection that have, at least one, of the multiples options that I pass for the query.
I was trying with the operator $elemMatch something like this:
db.collectioName.find({"options.name": { $elemMatch: {"optName1","optName2"}}})
but it never show me the matches documents.
Can someone help and show me, what I'm doing wrong?
Thanks!
Given a collection which contains the following documents:
{
"_id" : ObjectId("5a023b8d027b5bd06add627a"),
"name" : "valueName",
"options" : [
{
"idOption" : "optionId",
"name" : "optName1"
},
{
"idOption" : "optionId",
"name" : "optName2"
}
]
}
{
"_id" : ObjectId("5a023b9e027b5bd06add627d"),
"name" : "valueName",
"options" : [
{
"idOption" : "optionId",
"name" : "optName3"
},
{
"idOption" : "optionId",
"name" : "optName4"
}
]
}
This query ...
db.collection.find({"options": { $elemMatch: {"name": {"$in": ["optName1"]}}}})
.. will return the first document only.
While, this query ...
db.collection.find({"options": { $elemMatch: {"name": {"$in": ["optName1", "optName3"]}}}})
...will return both documents.
The second example (I think) meeets this requirement:
I'm trying to get all the documents in the collection that have, at least one, of the multiples options that I pass for the query.

How to Avoid Duplicate Entries in MongoDb Meteor App

How to avoid duplicate entries in mongoDb in Meteor application.
On the command: db.products.find({},{"TEMPLATE_NAME": 1},{unique : true})
{ "_id" : ObjectId("5555d0a16ce3b01bb759a771"), "TEMPLATE_NAME" : "B" }
{ "_id" : ObjectId("5555d0b46ce3b01bb759a772"), "TEMPLATE_NAME" : "A" }
{ "_id" : ObjectId("5555d0c86ce3b01bb759a773"), "TEMPLATE_NAME" : "C" }
{ "_id" : ObjectId("5555d0f86ce3b01bb759a774"), "TEMPLATE_NAME" : "C" }
{ "_id" : ObjectId("5555d1026ce3b01bb759a775"), "TEMPLATE_NAME" : "A" }
{ "_id" : ObjectId("5555d1086ce3b01bb759a776"), "TEMPLATE_NAME" : "B" }
I want to retrieve only the unique template names and show them on HTML page.
Use the aggregation framework where your pipeline stages consist of the $group and $project operators respectively. The $group operator step groups the input documents by the given key and thus will return distinct documents in the result. The $project operator then reshapes each document in the stream, such as by adding new fields or removing existing fields:
db.products.aggregate([
{
"$group": {
"_id": "$TEMPLATE_NAME"
}
},
{
"$project": {
"_id": 0,
"TEMPLATE_NAME": "$_id"
}
}
])
Result:
/* 0 */
{
"result" : [
{
"TEMPLATE_NAME" : "C"
},
{
"TEMPLATE_NAME" : "A"
},
{
"TEMPLATE_NAME" : "B"
}
],
"ok" : 1
}
You could then use the meteorhacks:aggregate package to implement the aggregation in Meteor:
Add to your app with
meteor add meteorhacks:aggregate
Then simply use .aggregate function like below.
var products = new Mongo.Collection('products');
var pipeline = [
{
"$group": {
"_id": "$TEMPLATE_NAME"
}
},
{
"$project": {
"_id": 0,
"TEMPLATE_NAME": "$_id"
}
}
];
var result = products.aggregate(pipeline);
-- UPDATE --
An alternative that doesn't use aggregation is using underscore's methods to return distinct field values from the collection's find method as follows:
var distinctTemplateNames = _.uniq(Collection.find({}, {
sort: {"TEMPLATE_NAME": 1}, fields: {"TEMPLATE_NAME": true}
}).fetch().map(function(x) {
return x.TEMPLATE_NAME;
}), true)
;
This will return an array with distinct product template names ["A", "B", "C"]
You can check out some tutorials which explain the above approach in detail: Get unique values from a collection in Meteor and METEOR – DISTINCT MONGODB QUERY.
You can use distinct of mongodb like :
db.collectionName.distinct("TEMPLATE_NAME")
This query will return you array of distinct TEMPLATE_NAME