Mongodb find query based on an array field

Question

The collection:

{
         "_id" : ObjectId("57506d74c469888f0d631be6"),
         "name" : "mycollection",
         "details" : [ 
             {
                 "date" : "25/03/2020",
                 "number" : "A",
                 "active" : false
              }
        },
{
        "_id" : ObjectId("57506d74c469888f0d631usi"),
        "name" : "newcollection",
        "details" : [ 

            {
                "date" : "30/03/2020",
                "number" : "C",
                "active" : false
            } 
        },
{
        "_id" : ObjectId("57506d74c4633388f0d631usi"),
        "name" : "mycollection",
        "details" : [ 

            {
                "date" : "31/03/2020",
                "number" : "C",
                "active" : false
            }
        },
    }

The find query to get the values based on the active status inside the details field.

I have tried:

db.collection.find(
    {"name": "mycollection", "details": {"active": False}})

Expected result: I need the collections where the active is false under details field in each collection. For here record id ObjectId("57506d74c469888f0d631be6") and ObjectId("57506d74c4633388f0d631usi") should be displayed.

above query seems good to me or you can try this db.collection.find( {"name": "mycollection", "details.active": false}}) // case sensitive of boolean value — harshit kohli
– harshit kohli, Commented May 27, 2020 at 14:30
like what if details array also contains some objects with active true and some with false you still require that document?@Navi — harshit kohli
– harshit kohli, Commented May 27, 2020 at 14:49

Mohammed Yousry · Accepted Answer · 2020-05-27 15:50:19Z

2

If details array may have some objects with active = true, and some objects = false, and you need to get only the objects that have active = false, then we can use $filter in aggregation pipeline to filter the details array

your query maybe something like that

db.collection.aggregate([
  {
    $match: {
      name: "mycollection",
      "details.active": false
    }
  },
  {
    $project: {
      name: 1,
      details: {
        $filter: {
          input: "$details",
          as: "elem",
          cond: {
            $eq: ["$$elem.active", false]
          }
        }
      }
    }
  }
])

you can test it here

hope it helps

answered May 27, 2020 at 15:50

Mohammed Yousry

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4 Comments

Navi Over a year ago

Can you post it like group by active?

Mohammed Yousry Over a year ago

something like this? mongoplayground.net/p/77W3WbjBPqu

Navi Over a year ago

No, it is not @Mohammed

Mohammed Yousry Over a year ago

Could you explain how do you mean by group by active?

Joseph Ireland · Accepted Answer · 2020-05-27 14:38:12Z

1

The docs contain a section on this type of query here: https://docs.mongodb.com/manual/tutorial/query-array-of-documents/

Your query doesn't work because that syntax requires the array to contain that exact object, but documents in your array contain extra fields, so none match.

The query for this case is db.collection.find({"name": "mycollection", "details.active": False}).

Note: this will return all documents where the array contains objects with active==false, but it won't filter the actual arrays to remove any elements with active=true.

answered May 27, 2020 at 14:38

Joseph Ireland

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1 Comment

Navi Over a year ago

Still no clue@Joseph

Santhosh · Accepted Answer · 2020-05-27 15:10:01Z

0

The query for mongodb or robomongo is the same as in your case the find() might work, but I choose aggregate which could filter the values.

Example query:

db.getCollection('collection').aggregate([{$match: {'details.active': {$ne: true},"name":"mycollection"}}])

Simple. But kindly verify and revert.

answered May 27, 2020 at 15:10

Santhosh

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