Question

I try to manipulate a variable inside a function. But it seems to forget the values once I exit the function, eventhough the variable is declared outside the function.

The essential code:

var posts = {};                

            // Perform a data request
            // skjutsgruppens-page
            $.oajax({
                url: "https://graph.facebook.com/197214710347172/feed?limit=500",
                    *SNIP*
                    success: function(data) {
                    $.extend(posts, data);
                }
            });


            // Gruppen
            $.oajax({
                url: "https://graph.facebook.com/2388163605/feed?limit=500",
                *snip*
                success: function(data) {
                    $.extend(posts, data);
                }
            });

The oajax retrievies data from facebook. I want to make a variable that contains the data from both oajax methods.

The actual code: http://eco.nolgren.se/demo/resihop/#

Answer 1

The issue is likely that the success function executes at an arbitrary time in the future–unless you specifically access posts after you know the success function has executed, you will receive undefined results, completely dependent on function and access timing.

The best approach is to handle this correctly by doing necessary work inside in the success function, or use something like jQuery’s .when function.