views:

416

answers:

2
$.getJSON('http://twitter.com/followers/ids.json?screen_name=' + query1 + '&callback=?', function (data) {
    console.log('JSON data string 1 is: ' + data);
    $.getJSON('http://twitter.com/followers/ids.json?screen_name=' + query2 + '&callback=?', function (data1) {
        console.log('JSON data string 2 is: ' + data1);
        f2 = data1;
        f1 = data;
        for (var i = 0; i < f1.length; i++) {
            for (var j = 0; j < f2.length; j++) {
                if (f1[i] == f2[j]) { //console.log("Adding f1[i]");
                    common.push(f1[i]);
                }
            }
        }
        for (var d = 0; d < common.length; d++) {
            $.getJSON('http://twitter.com/users/show.xml?user_id=' + common[d] + '&callback=?', function (data2) {
                $('#content').append('<>img  width="50" height="50" src=' + data2.profile_image_url + '>< 

href="http://www.twitter.com/' + data2.screen_name + '">' + data2.name + '</></>');
            });
        }
    });
});

in this code basically im getting an array(common[]) that contains all the common followers between two twitter users

but the third $.getJSON call doesnt seem to responding at all

Am i doing something wrong

Any help will be appreciated

Thank You

+2  A: 

setup a global AJAX error event handler like so, and you can then inspect the error. remember if an error occurs within the context of a jQuery AJAX request then nothing happens outside of the error function. if you are using the $.getJSON method then you must set up the error handler in the $.ajaxSetup method.

$.ajaxSetup({
    error: function(xhr, status, e) {
        console.log(xhr, status, e);
    }
});
Jon Erickson
A: 

got it working...the problem was that the url had show.xml ...i just changed it to show.json and it started working

anand
not sure if you are using FireBug, but you definately should be using it to inspect your DOM and scripts. makes finding these things much easier.
Jon Erickson
thanks..il use firebug then...
anand