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Variable doesn't get returned from AJAX function

As my framework grows i decided to split it into files, instead of leaving it in the main design file. However by doing that the return of a function doesn't return any value.

data isn't empty - if i alert the values in the js file they are there!

The functions:

1st the function in .js file (is included before the execution)

             var lock_get = 0;
             function get_data(data, destination) 
             {

                if (lock_get == 0)
                {
                    lock_get = 1;
                    $.ajax({
                        type: "POST",
                        url: destination,
                        async: true,
                        data: data,
                        success: function(data) 
                        {
                            lock_get = 0;
                            if (data)
                            {
                                return data;
                            }
                        }
                    });
                }
             };

So and here is the execution part:

    var test = get_data(data, destination);
    notice(test);

and test is empty... I already tried different ways for writing but I guess i missunderstood the possibilities of js?

like image 380
Ivan Schrecklich Avatar asked Sep 18 '12 10:09

Ivan Schrecklich


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1 Answers

You can't do that : as the call is asynchronous, the get_data function can't return the result of the ajax call.

What you should do is provide a callback to the get_data function and handle the result in the callback.

function get_data(data, destination, callback) 
         {

            if (lock_get == 0)
            {
                lock_get = 1;
                $.ajax({
                    type: "POST",
                    url: destination,
                    async: true,
                    data: data,
                    success: function(data) 
                    {
                        lock_get = 0;
                        if (data && callback)
                        {
                            callback(data);
                        }
                    }
                });
            }
         };

And call it like that :

get_data(data, destination, function(test){
   notice(test);
});
like image 176
Denys Séguret Avatar answered Sep 19 '22 13:09

Denys Séguret