CODEWITHSUNDEEP
mongodbmongodb-queryaggregation-framework
invzbl3
invzbl3

Reputation: 6520

How to project fields in mongoDB?

I have the structure of documents in the collection like this:

{
  "_id" : ObjectId("569190cd24de1e0ce2dfcd62"),
  "title" : "Star Trek II: The Wrath of Khan",
  "year" : 1982,
  "rated" : "PG",
  "released" : ISODate("1982-06-04T04:00:00Z"),
  "runtime" : 113,
  "countries" : [
    "USA"
  ],
  "awards" : {
    "wins" : 2,
    "nominations" : 9,
    "text" : "2 wins & 9 nominations."
  }
}

I'm trying to get the contents of specific fields using the projection and adding a couple of additional parameters. I want to take such keys as: title, year, rated and awards with the specified values. (and _id remove)

I write so db.movieDetails.find( {}, {title: 1, year: 2013, rated: "PG-13", _id: 0, "awards.wins": 1 }).pretty() to get the fields with values, but console displays different parameters:

{
   "title" : "Star Trek II: The Wrath of Khan",
   "year" : 1982,
   "rated" : "PG",
   "awards" : {
     "wins" : 2,
   }
}

I want to have the output like this, for example:

{
   "title" : "Star Trek II: The Wrath of Khan",
   "year" : 2013,
   "rated" : "PG-13",
   "awards" : {
     "wins" : 0
   }
}

Tell me, please, what needs to be corrected in the query to have only my concrete requirements... I appreciate everyone's help.

Upvotes: 5

Views: 4491

Answers (2)

Serdar
Serdar

Reputation: 478

Filtering should be done inside first parameter of the find() function and projection should be done inside second parameter of the find function.

Upvotes: 2

Ashh
Ashh

Reputation: 46491

You can do something like that

db.collection.aggregate([
  {
    $project: {
      title: 1,
      _id: 0,
      year: 2013,
      rated: "PG-13",
      "awards.wins": "0"
    }
  }
])

Will give following output

[
  {
    "awards": {
      "wins": "0"
    },
    "rated": "PG-13",
    "title": "Star Trek II: The Wrath of Khan",
    "year": 1982
  }
]

Upvotes: 5

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