Wrap all fields of MongoDB Document into a root/top level object

Question

Suppose I have two documents in a collection:

[
  {
    "_id": 0,
    "name": "aaaa",
    "phone": 111
  },
  {
    "_id": 1,
    "name": "bbbb",
    "phone": 999
  }
]

I want to run an aggregation pipeline that will merge/wrap/combine ALL the fields into a top-level root object called lead. Without mentioning/hardcoding the field names in the query.

Expected output:

[
  {
    "lead": {
      "_id": 0,
      "name": "aaaa",
      "phone": 111
    }
  },
  {
    "lead": {
      "_id": 1,
      "name": "bbbb",
      "phone": 999
    }
  }
]

score 2 · Accepted Answer · answered Apr 06 '23 at 01:09

2

Works with $replaceWith stage to replace the input documents with new outputdocuments. To get each (original) document, you need to use the $$ROOT variable.

db.collection.aggregate([
  {
    $replaceWith: {
      "lead": "$$ROOT"
    }
  }
])

Demo @ Mongo Playground

answered Apr 06 '23 at 01:09

Yong Shun

35,286
4
24
46

Can also be achieved with `$project` but I'm not sure which one is efficient/recommend? https://stackoverflow.com/a/75945067/9486457 – Dürrani Apr 06 '23 at 01:11
1

From my perspective, I think the performance is not much different. This `$replaceWith` is an alternative to `$project` and consider as a *shortcut*. – Yong Shun Apr 06 '23 at 02:53

score 0 · Answer 2 · answered Apr 06 '23 at 01:06

0

Got some help from this answer and was able to achieve it using $project:

db.collection.aggregate([
  {
    $project: {
      lead: "$$ROOT",
      _id: false
    }
  }
])

Playground link

$$ROOT in MongoDB references the content of this document.

answered Apr 06 '23 at 01:06

Dürrani

950
12
21

Wrap all fields of MongoDB Document into a root/top level object

2 Answers2