MongoDB Join Query - localField Using _id for Matching Not Working

Scenario

Here's the situation - I have two documents: Document A

_id	name
6090acbd682e000028007e74	jack
6090acbd682e000028007e75	rose

Document B

_id	a_id	state
6787848807839043584	6090acbd682e000028007e74	1
6787848807839043585	6090acbd682e000028007e75	0

Then I need to join query Document A and Document B.

In MySQL we can do:

select * from A left join B on A._id=B.a_id

But in MongoDB we need to implement this through aggregate. Initially I wrote:

db.A.aggregate([
{$lookup:{
	from: "B",
         localField: "_id",
         foreignField: "a_id",
         as: "success_id"

}}])

Found there were no associated values at all.

Solution

Found a similar question on Stack Overflow. Turns out you need to add this to the query:

{
    $addFields: { "_id": { "$toString": "$_id" } }
  },

So the query becomes:

db.A.aggregate([
{
    $addFields: { "_id": { "$toString": "$_id" } }
  },
{$lookup:{
from: "B",
         localField: "_id",
         foreignField: "a_id",
         as: "success_id"

}}])

This gives the joined result.

Scenario​

Solution​

Scenario

Solution