Mongodb Join on _id field from String to ObjectId

53,105

Solution 1

This is not possible as of MongoDB 3.4. This feature has already been requested, but hasn't been implemented yet. Here are the corresponding tickets:

For now you'll have to store userId as ObjectId


EDIT

The previous tickets were fixed in MongoDB 4.0. You can now achieve this with the folowing query:

db.user.aggregate([
  {
    "$project": {
      "_id": {
        "$toString": "$_id"
      }
    }
  },
  {
    "$lookup": {
      "from": "role",
      "localField": "_id",
      "foreignField": "userId",
      "as": "role"
    }
  }
])

result:

[
  {
    "_id": "584aac38686860d502929b8b",
    "role": [
      {
        "_id": ObjectId("584aaca6686860d502929b8d"),
        "role": "Admin",
        "userId": "584aac38686860d502929b8b"
      }
    ]
  }
]

try it online: mongoplayground.net/p/JoLPVIb1OLS

Solution 2

You can use $toObjectId aggregation from mongodb 4.0 which converts String id to ObjectId

db.role.aggregate([
  { "$lookup": {
    "from": "user",
    "let": { "userId": "$_id" },
    "pipeline": [
      { "$addFields": { "userId": { "$toObjectId": "$userId" }}},
      { "$match": { "$expr": { "$eq": [ "$userId", "$$userId" ] } } }
    ],
    "as": "output"
  }}
])

Or you can use $toString aggregation from mongodb 4.0 which converts ObjectId to String

db.role.aggregate([
  { "$addFields": { "userId": { "$toString": "$_id" }}},
  { "$lookup": {
    "from": "user",
    "localField": "userId",
    "foreignField": "userId",
    "as": "output"
  }}
])

Solution 3

I think the previous answer has an error on the $toObjectId case. The let statement applies to the db collection on which the function aggregate is called (i.e 'role') and not on the collection pointed by "from" (i.e 'user').

db.role.aggregate([
  { "$lookup": {
    "let": { "userObjId": { "$toObjectId": "$userId" } },
    "from": "user",
    "pipeline": [
      { "$match": { "$expr": { "$eq": [ "$_id", "$$userObjId" ] } } }
    ],
    "as": "userDetails"
  }}
])

Or

db.role.aggregate([
  { "$project": { "userObjId": { "$toObjectId": "$userId" } } },
  { "$lookup": {
    "localField": "userObjId",
    "from": "user",
    "foreignField": "$_id",
    "as": "userDetails"
  }}
])

And

db.user.aggregate([
  { "$project": { "userStrId": { "$toString": "$_id" }}},
  { "$lookup": {
    "localField": "userStrId",
    "from": "role",
    "foreignField": "userId",
    "as": "roleDetails"
  }}
])
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53,105
Kavya Mugali
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Kavya Mugali

Developer working on C# .Net, MVC, WCF and Node js. Learning AngularJs, Software Design, Application Security Keen on knowing and following best practices and industry standards.

Updated on July 09, 2022

Comments

  • Kavya Mugali
    Kavya Mugali almost 2 years

    I have two collections

    1. User

      {
         "_id" : ObjectId("584aac38686860d502929b8b"),
         "name" : "John"
      }
      
    2. Role

      {
         "_id" : ObjectId("584aaca6686860d502929b8d"),
         "role" : "Admin",
         "userId" : "584aac38686860d502929b8b"  
      }
      

    I want to join these collection based on the userId (in role collection) - _id ( in user collection).

    I tried the below query:

    db.role.aggregate({
      "$lookup": {
        "from": "user",
        "localField": "userId",
        "foreignField": "_id",
        "as": "output"
      }
    })
    

    This gives me expected results as long as i store userId as a ObjectId. When my userId is a string there are no results. Ps: I tried

    foreignField: '_id'.valueOf()

    and

    foreignField: '_id'.toString()

    . But no luck to match/join based on a ObjectId-string fields.

    Any help will be appreciated.

  • Nech
    Nech over 5 years
    Note that these 2 aggregation operators are available only from 4.0
  • AJB
    AJB over 5 years
    @AnthonyWinzlet This should now be the accepted answer.
  • felix
    felix about 4 years
    Hi @Ashh, you have the wrong bracket in cond, this should work: mongoplayground.net/p/jKVt60tPtc6
  • teuber789
    teuber789 almost 3 years
    Good callout on the error in the previous answer's let statement.
  • teuber789
    teuber789 almost 3 years
    There's an error in the $toObjectId case. Gary Wild's answer fixes it.