How to remove a document in Mongoose, only if one of its fields matches some condition?

Question

I want to delete one item only if it matches some condition. It would be something like this:

Session.findOne({ id:req.body.sessionId },function(err,session){
    if(session.user == req.body.userId) {
        session.remove();
    }
});

Which obviously doesn't work. How can I achieve that?

Answer 1

You can test multiple conditions at the same time in Mongoose. This should work for you.

Session.findOne({ 
  id: req.body.sessionId,
  user: req.body.userId
})
.remove()
.exec();

Answer 2

Can you not just combine the query?

Session.findOne({ id:req.body.sessionId, user: req.body.userId}).remove( callback );

Answer 3

Honestly, just do :

Session.remove({ "_id": req.body.sessionId },function(err,result) {

});

That's more efficient and safe.

The reason it's failing in your example is because the variable is not a string anymore but an ObjectId in the returned document. But it's not really safe so don't do it that way.

---

If you are being paranoid about the matching session then do it first:

if (session.user.toString() == req.body.userId) {
    Session.remove({ "_id": req.body.sessionId },function(err,result) {

    });
} else {
   // Not the same
}

And be aware that an ObjectId is not the same as a string in the request. Convert or cast one of them.