How to convert Mongo query array result into flatten object and key should be _id

Question

How to convert mongo query result into flatten object.

Current result:

[
  {
    "_id": "61ef7e5497c9912ac1e56404",
    "name": "Mark",
    "age": 20,
    "isMinor": true
  },
  {
    "_id": "61ef7e5497c9912ac1e56404",
    "name": "Sam",
    "age": 22,
    "isMinor": false
  }
]

Excepted output:

[
  {
    "61ef7e5497c9912ac1e56404": {
      "_id": "61ef7e5497c9912ac1e56404",
      "name": "Mark",
      "age": 20,
      "isMinor": true
    },
    "61ef7e5497c9912ac1e56404": {
      "_id": "61ef7e5497c9912ac1e56404",
      "name": "Sam",
      "age": 22,
      "isMinor": false
    }
  }
]

I wanted to do this using aggregation query itself.

My current query:

Users.aggregate({$addField:{isMinor:{ $cond: [
                    { $gt: ['$age', 20] },
                    true, false,
                  ]}}})

Yong Shun · Accepted Answer · 2022-01-25 11:59:11Z

1

$group - Group the document and add the users field which holds the key-value pair (key: id, value: document) array. This stage will return only a single document.
$replaceWith - Replaces the input document with the specified document. This stage will output the document in key-value pair.

db.collection.aggregate([
  // Previous stages
  {
    $group: {
      _id: null,
      users: {
        $push: {
          k: { $toString: "$_id" },
          v: "$$ROOT"
        }
      }
    }
  },
  {
    "$replaceWith": {
      $arrayToObject: "$users"
    }
  }
])

Sample Mongo Playground

edited Jan 25, 2022 at 11:59

answered Jan 25, 2022 at 10:09

Yong Shun

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2 Comments

Cool_Ashish Over a year ago

Thanks, @yong. There is only one small change for real-time execution. k: { $toString: '$_id' }. _id needs to convert into a string as arrayToObject does not convert objected type value to key.

Yong Shun Over a year ago

Ya, key must be in string. You're welcome. =)

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