Jquery AJAX response JSON in variable.

Question

I have two functions. The first sets a variable then the second get's the information needed for the first function. The issue I was first having was that everything was getting returned before the ajax response was happening. Basically what I need to do is get a JSON object and set it to a variable, then uses certain aspects of that object and append them to items. here's what I have...

$(document).ready(function(){

                $('#youtube-url').blur(function() {

                 $info = youtubeInfo($('#youtube-url').val());

                 console.log($info);
                 if($info.html){
                    //alert($info.thumbnail_url);
                    //$('#preview_video_thumb').attr('src', $info.thumbnail_url);
                 }


                });


            });
            function youtubeInfo(url){
                var odata = '';
                var result = jQuery.ajax({
                          url: '<? echo base_url("videos/getOembed") ?>',
                          type: 'POST',
                          dataType: 'json',
                          async: false,
                          data: {url: url},
                          complete: function(xhr, textStatus) {

                          },
                          success: function(data, textStatus, xhr) {
                           return data;
                          },
                          error: function(xhr, textStatus, errorThrown) {
                            //called when there is an error
                          }
                        }).responseText;
                return result;

            };

If I return the 'data' object from within the success callback it returns blank. if I do it how it is, the responseText of the AJAX call is a string and not JSON. any ideas? I basically need to set the $info variable in the first function to the AJAX response (JSON) of the second function.

user400654 · Accepted Answer · 2012-05-29 20:01:04Z

3

Remove .responseText and use this:

$('#youtube-url').blur(function() {
    youtubeInfo($('#youtube-url').val()).done(function($info){
        if($info.html){
            //alert($info.thumbnail_url);
            //$('#preview_video_thumb').attr('src', $info.thumbnail_url);
        }
    });
});

answered May 29, 2012 at 20:01

user400654

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3 Comments

G.Thompson Over a year ago

Wow man, that's exactly it! Do you mind explaining why that works? I never even seen the done function before.

user400654 Over a year ago

It works because jQuery ajax methods implement the Deferred object interface which includes four methods: done, fail, always, and then. There are a few others, but these four are the most important. done and fail are obvious, always will always be called (obvious too i guess) and then allows you to define both done and fail at the same time .then(donehandler,failhandler)

user400654 Over a year ago

@GThompson -- you can also remove async: false if you are using .done()

Matt Dodge · Accepted Answer · 2012-05-29 20:02:34Z

0

The return data statement you have in the success function is not doing what you intend. It is not actually returning out of the ajax call, but rather returning out of the success function. What you probably want is something like this instead:

function youtubeInfo(url){
            var odata = '';
            var result = jQuery.ajax({
                      url: '<? echo base_url("videos/getOembed") ?>',
                      type: 'POST',
                      dataType: 'json',
                      async: false,
                      data: {url: url},
                      complete: function(xhr, textStatus) {

                      },
                      success: function(data, textStatus, xhr) {
                       odata = data;
                      },
                      error: function(xhr, textStatus, errorThrown) {
                        //called when there is an error
                      }
                    });
            return odata;

        };

answered May 29, 2012 at 20:02

Matt Dodge

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3 Comments

Evan Davis Over a year ago

This will still return odata before it is set.

Matt Dodge Over a year ago

It shouldn't because the OP is setting async to false

user400654 Over a year ago

@Mathletics no, it will return after it is set, async:false

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