How to get $oid hash value as string with mongo aggregate framework?

Question

I have a collection with objects with a json output like this:

{
  "_id": {
    "$oid": "53107ed2e6fb449fa641b903"
  },
  "title": "myBook"
}

But I would like to get this:

{
  "_id": "53107ed2e6fb449fa641b903",
  "title": "myBook"
}

I have tried with dot notation and $project but it doesn't work:

db.books.aggregate({
  $project: {
    _id: "$_id.oid",
    title: 1 
  }
)

possible duplicate of [Ruby model output id as object oid](http://stackoverflow.com/questions/18646223/ruby-model-output-id-as-object-oid) — mu is too short, Mar 01 '14 at 19:35

score 1 · Answer 1 · answered Mar 02 '14 at 00:02

1

Looks like your aggregate query syntax is incorrect. Try this:

db.books.aggregate([{$project:{_id:"$_id.oid", title:1}}])

Also, in the input json, you have a '$' prefix for the oid (i.e, "$oid"). That's illegal in MongoDB.

answered Mar 02 '14 at 00:02

Anand Jayabalan

I that is the whole point of his question. How to do this with an $oid. – A.com Dec 27 '16 at 00:21
https://stackoverflow.com/questions/36059986/how-to-project-objectid-to-string-value-in-mongodb-aggregate – Hariharan Kanakaraja Aug 12 '21 at 17:19

1 Answers1