Using two $.get in one. Trying to use as a variable

Question

This is my code

$(function () {
   $.get('/viewonline', function (data) {
    data = $(data);
    var members = data.find('.userdata');
   for (var j = 0; j < members.length; j++) {
     var membername = $(members[j]).find('.username').text();
     var memberURL = $(members[j]).find('.username').attr('href');
     var memberAva = $.get(memberURL, function(data) {
                    data = $(data);
                    data.find('#profile-advanced-right img:eq[0]').attr('src');
       });
  $('.user_info_on').append('<div class="on_name"><a href="' + memberURL + '" title="'+ membername +'"><img src="' + memberAva + '"/></a></div>');
       }
    }, 'html');
});

As you can see in the var memberAva area I use another $.get function. I am trying to use the users URL to pass this function then place it all together. Though I get [object Object] which honestly is what I figured. Is this possible to use two $.get together? If not what is my best option to grab the image src that I need?

Try for commenting callback- Like this?

  $(function () {
   $.get('/viewonline', function (data) {
    data = $(data);
    var members = data.find('.userdata');
   for (var j = 0; j < members.length; j++) {
     var membername = $(members[j]).find('.username').text();
     var memberURL = $(members[j]).find('.username').attr('href');
     var memberAva = $.get(memberURL, function(image) {
                    image = $(image);
                    image.find('#profile-advanced-right img:eq[0]').attr('src');
       });
  $('.user_info_on').append('<div class="on_name"><a href="' + memberURL + '" title="'+ membername +'"><img src="' + memberAva + '"/></a></div>');
       }
    }, 'html');
});

Welcome to the wonderful world of **async**! You need another callback. — SLaks, May 19 '13 at 03:04
Oh why thank you for welcoming me ;) Now what do you mean another call back? (data , image) ? — EasyBB, May 19 '13 at 03:06
ok so let me see if I have it right, I will update my first post with my try ok :) — EasyBB, May 19 '13 at 03:07
http://stackoverflow.com/questions/14220321/how-to-return-the-response-from-an-ajax-call — James Montagne, May 19 '13 at 03:42

score 1 · Answer 1 · answered May 19 '13 at 03:37

1

I think you need to create a global variable and when your second AJAX request succeeds you may set the value of that global variable and can use anywhere.

answered May 19 '13 at 03:37

Talha

11
1

Using two $.get in one. Trying to use as a variable

1 Answers1