How to set value of outer variable in anonymous function

Question

I am kind of a beginner to javascript. This is a javascript problem, but for the record, this is a program done in node.js using hapi and mongoose. Tables is a model in mongoose.

My problem is that I want to set the response to a value so I can send it later:

handler: function (request, reply) {
    var response;
    Tables.find(function find(err, mesas) {
        if (err) 
            response = reply({ error: err});
        else 
            response = reply({mesas : mesas}).hold;
        });
    }
    response.send();
}

For some reason, server returns "undefined does not have method 'send'", and I guess that is because it is referenced inside an anonymous function. This brings a couples of thoughts to me:

1) Am I writing this in a non javascript idiomatic way?
2) What is the best approach to to makes this work?

EDIT: The response.send() call was actually outside of the Tables.find(...).

Again the everyday async question... You can't return inside a continuation. Use promises, or do the logic inside the callback. — elclanrs, Jun 03 '14 at 23:50
I apologize if I asked a duplicate, please link to it if I did. — Guido Tarsia, Jun 03 '14 at 23:52
response.send(), will be executed before the Table.find callback. You have to send the response back to the client within the Table.find callback. — Robert Moskal, Jun 03 '14 at 23:54
No need to apologize, it's a common duplicate yes, but it's a common misunderstanding as well. — elclanrs, Jun 03 '14 at 23:54
@RobertMoskal I'm sorry, I edited it, response.send() is called outside. — Guido Tarsia, Jun 03 '14 at 23:57
@elclanrs Oh, thanks, that done()/fail() call at the end is nice indeed. — Guido Tarsia, Jun 04 '14 at 00:00
Keep in mind that syntax is probably library dependent. Not sure about mongoose, but it is probable that it returns a promise. — elclanrs, Jun 04 '14 at 00:05

How to set value of outer variable in anonymous function

0 Answers0