Arrgregate $out mongodb insert for $out failed duplicate Error

Question

Query1: i was trying to combine two collection with selected field of both collections using mongodb Aggregate $lookup as follows

 db.col1.aggregate([
   {
              


        

Answer 1

Ok for duplicated id explanation. In addition, if you want to keep user._id, add id:"$_id" to your last project stage.

But for the second query with _id:0, what do you mean by 'no error and no output :

• If your query is running well, results will be inserted in a new collection called 'new_col', so you have to run db.getCollection("new_col").find({}); to retrieve them. Nothing in console with $out stage!
• If your new_col collection doesn't exist or is empty : your query is returning nothing, no matter of $out stage. In this case, please provide col1 and col2 document samples

---EDIT--- With these data :

col1 : 
{ 
"_id" : 1.0, 
"field1" : "a", 
"field2" : "b"
}
col2 : 
{ 
"_id" : 1.0, 
"field1" : "a", 
"field2" : "c"
}

the right query to achieve what you need is the following :

db.col1.aggregate(
[
    // Stage 1
    {
        $lookup: {
             from: "col2",
                  localField: "field1",    
                  foreignField: "field1", 
                  as: "user"
        }
    },
    // Stage 2
    {
        $unwind: {
            path : "$user",

        }
    },
    // Stage 3
    {
        $project: {
           "userfield": "$user.field1",field1:1
        }
    },

]
);

Will output :

{ 
    "_id" : 1.0, 
    "field1" : "a", 
    "userfield" : "a"
}