Get distinct field names of sub documents in MongoDB

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Let’s say you have these documents in your collection items:

db.items.find()

{ "_id" : ObjectId("5f0345275663006139066197"), "subDocument" : { "field1" : 42 } }
{ "_id" : ObjectId("5f03452c5663006139066198"), "subDocument" : { "field3" : 6 } }
{ "_id" : ObjectId("5f0345275663006139066199"), "subDocument" : { "field1" : 6 } }

In other words the fields of the sub document subDocument are not the same. They could be user-defined, or simply because of the nature of the domain you’re working in.


So, how would you get the distinct field name of those sub-documents?

As a results I would like to have an array containing the different field names.

db.items.aggregate({
  $project: {
    subDocument: {
      $objectToArray: "$subDocument"
    }
  }
}, {
  $unwind: '$subDocument'
}, {
  $project: {
    _id: '$subDocument.k'
  }
},
{
  $group: {
    _id: '$_id'
  }
})

{ "_id" : "field1" }
{ "_id" : "field3" }

Now you can just map each document and extract the _id field to have the distinct field names of all sub-documents.

Explanatation

$project with $objectToArray

$objectToArray comes in handy in this case to destructure the object into [key, value] pairs in the following format:

db.items.aggregate({$project: { subDocument: { $objectToArray: "$subDocument" } }})
{ "_id" : ObjectId("5f0345275663006139066197"), "subDocument" : [ { "k" : "field1", "v" : 42 } ] }
{ "_id" : ObjectId("5f03452c5663006139066198"), "subDocument" : [ { "k" : "field3", "v" : 6 } ] }
{ "_id" : ObjectId("5f0346c85663006139066199"), "subDocument" : [ { "k" : "field1", "v" : 6 } ] }

$unwind the subDocument array

We want to have objects to get the fields, so you unwind (kind of “unzip”) the array in distinct objects.

db.items.aggregate({$project: { subDocument: { $objectToArray: "$subDocument" } }}, {$unwind: '$subDocument'})
{ "_id" : ObjectId("5f0345275663006139066197"), "subDocument" : { "k" : "field1", "v" : 42 } }
{ "_id" : ObjectId("5f03452c5663006139066198"), "subDocument" : { "k" : "field3", "v" : 6 } }
{ "_id" : ObjectId("5f0346c85663006139066199"), "subDocument" : { "k" : "field1", "v" : 6 } }

$project just the k field

We are interested in each k (key) field of the subDocuments (that now are objects, instead of arrays after the $unwind stage):

db.items.aggregate({$project: { subDocument: { $objectToArray: "$subDocument" } }}, {$unwind: '$subDocument'}, {$project: {_id: '$subDocument.k'}})
{ "_id" : "field1" }
{ "_id" : "field3" }
{ "_id" : "field1" }
``` 

## $group by _id to get rid of duplicate fields

`$group` can be used similar to `.distinct`, but in an aggregation phase.

In this case, we want to "group" by the name of the fields, so that we have unique values:

```js
db.items.aggregate({$project: { subDocument: { $objectToArray: "$subDocument" } }}, {$unwind: '$subDocument'}, {$project: {_id: '$subDocument.k'}}, {$group: {_id: '$_id'}})
{ "_id" : "field1" }
{ "_id" : "field3" }

Now you can map the _id fields and extract the values, to finally have the unique field names:

> groupedFields = db.items.aggregate({$project: { subDocument: { $objectToArray: "$subDocument" } }}, {$unwind: '$subDocument'}, {$project: {_id: '$subDocument.k'}}, {$group: {_id: '$_id'}}).toArray()
[ { "_id" : "field3" }, { "_id" : "field1" } ]
> groupedFields.map(function (g) {return  g._id})
[ "field3", "field1" ]

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