MongoDB Bulk Insert Ignore Duplicate

Question

I\'ve Googled around and can\'t find any solid information on how to ignore duplicate errors when using bulk insert.

Here\'s the code I\'m currently using:

Answer 1

An alternative is to use bulk.find().upsert().replaceOne() instead:

MongoClient.connect(mongoURL, function(err, db) {
    if(err) console.err(err)
    let col = db.collection('user_ids')
    let batch = col.initializeUnorderedBulkOp()

    ids.forEach(function(id) {        
        batch.find({ userid: id }).upsert().replaceOne({ 
            userid: id, 
            used: false,  
            group: argv.groupID 
        });
    });

    batch.execute(function(err, result) {
        if(err) {
            console.error(new Error(err))
            db.close()
        }

        // Do some work

        db.close()
    });
});

With the above, if a document matches the query { userid: id } it will be replaced with the new document, otherwise it will be created hence there are No duplicate key errors thrown.

---

For MongoDB server versions 3.2+, use bulkWrite as:

MongoClient.connect(mongoURL, function(err, db) {

    if(err) console.err(err)

    let col = db.collection('user_ids')
    let ops = []
    let counter = 0

    ids.forEach(function(id) {
        ops.push({
            "replaceOne": {
                "filter": { "userid": id },
                "replacement": { 
                    userid: id, 
                    used: false,  
                    group: argv.groupID 
                },
                "upsert": true
            }
        })

        counter++

        if (counter % 500 === 0) {
            col.bulkWrite(ops, function(err, r) {
                // do something with result
                db.close()
            })
            ops = []
        }
    })

    if (counter % 500 !== 0) {
        col.bulkWrite(ops, function(err, r) {
            // do something with result
            db.close()
        }
    } 
})