CODEWITHSUNDEEP
mongodb
leiyonglin
leiyonglin

Reputation: 6814

MongoDB order by array children element?

This is my data:

 {
  "112233": [
    {
      "pos_opts": {
        "acc": 10,
        "time": "123456"
      }
    },
    {
      "pos_opts": {
        "acc": 20,
        "time": "1234567"
      }
    },
    {
      "pos_opts": {
        "acc": 20,
        "time": "1234568"
      }
    },
    {
      "pos_opts": {
        "acc": 30,
        "time": "1234569"
      }
    }
  ],
  "223344": [
    {
      "pos_opts": {
        "acc": 10,
        "time": "123456"
      }
    },
    {
      "pos_opts": {
        "acc": 20,
        "time": "1234567"
      }
    }
  ],
  "_id": "75172"
}

I want to get the result:

SELECT 112233.children FROM x WHERE _id=75172 ORDER BY pos_opts.time DESC LIMIT 2

the result like:

[
    {
      "pos_opts": {
        "acc": 30,
        "time": "1234569"
      }
    },
    {
      "pos_opts": {
        "acc": 20,
        "time": "1234568"
      }
    }
 ]

This is my code (of course doesn't works):

db.location.find({_id:"75172"}, {"112233": true}).sort({'pos_opts.time': -1})

Thanks.

Upvotes: 1

Views: 472

Answers (1)

Parvin Gasimzade
Parvin Gasimzade

Reputation: 26012

You can do it with Aggregation Framework. Query will look like : 

db.location.aggregate(
{$match : {_id:"75172"}},
{$project : {"112233" : 1}},
{$unwind : "$112233"},
{$sort : {"112233.pos_opts.time" : -1}},
{$limit : 2}
)

Result will look like : 

{
    "_id" : "75172",
    "112233" : {
        "pos_opts" : {
            "acc" : 30,
            "time" : "1234569"
        }
    }
},
{
    "_id" : "75172",
    "112233" : {
        "pos_opts" : {
            "acc" : 20,
            "time" : "1234568"
        }
    }
}

Upvotes: 2

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