Question 239830 in Mongodb Question posted in
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Mongodb – Combining Mongo query results into a single document

I have a Mongo collection as such:

db.items.find({}):

[{
  "_id": {
    "$oid": "64e929386a2bff904d86118a"
  },
  "name": "Copies",
  "legacy_id": 181
},
{
  "_id": {
    "$oid": "64e929386a2bff904d86118b"
  },
  "name": "Temp",
  "legacy_id": 182
},
{
  "_id": {
    "$oid": "64e929386a2bff904d86118c"
  },
  "name": "Test",
  "legacy_id": 183
}]

Is there a way to combine the data to have a single document that contains the key/value pairs of the "_id": "legacy_id" of the returned query results? A result like below would work fine for me. I know I can do this after the fact through some code but I’m wondering if there is a way to do this in the aggregate pipeline. Here’s the end result that I would like to achieve:

[{
  "64e929386a2bff904d86118a": 181,
  "64e929386a2bff904d86118b": 182,
  "64e929386a2bff904d86118c": 183,
}]
Tags:

2

Answers


  1. 0

    Perform unconditional $group to $push all k-v tuples of records into an array first. Then use $arrayToObject to convert the array into object.

    db.collection.aggregate([
  {
    "$group": {
      "_id": null,
      "records": {
        "$push": {
          k: {
            "$toString": "$_id"
          },
          v: "$legacy_id"
        }
      }
    }
  },
  {
    "$replaceRoot": {
      "newRoot": {
        "$arrayToObject": "$records"
      }
    }
  }
])

    Mongo Playground

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  2. 0

    you can use $arrayToObject and using $replaceRoot you can make it root

    test it here: mongo playgroung

    db.collection.aggregate([
  {
    $replaceRoot: {
      newRoot: {
        $arrayToObject: [
          [
            {
              k: "$_id",
              v: "$legacy_id"
            }
          ]
        ]
      }
    }
  }
])
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