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s.success is not a function

Submitted by Gold on Thu, 05/21/2009 - 17:31.

This post is about JavaScript, jQuery and annoying errors that don't seem to have a solution out there. Right, keywords done. Here's the problem and the solution I used;

I was having an issue with a jQuery call that was returning the following error;

s.success is not a function

The code that was generating this was the callback function on the successful return from a getJSON() call. This is the original script;

call = '/admin/ts2/pending/approve/'+sid+'/'+c1+'/'+c2+'/'+c3;
$.getJSON(call, '', 'process_approved');

After a decent amount of hunting I drew a blank. There's a reasonable number of queries about this but no answer that I could use. One was a patch on the original jQuery source but given I'm working in a framework I'm not able to touch that.

I eventually figured out a work around though. Instead of using a callback function I used an in-line function and from that called my external function. This is the code as it stands now;

call = '/admin/ts2/pending/approve/'+sid+'/'+c1+'/'+c2+'/'+c3;
$.getJSON(call, '', function(data)
  {
    process_approved(data);
  }
);

Hopefully this will be of use to someone.

»

Looks like you're trying to

Submitted by Anonymous on Mon, 12/21/2009 - 21:44.

Looks like you're trying to pass in a string instead of a function variable. In that case, s.success will be set to a string, not a function.

Instead, try taking off the quotation marks around 'process_approved':

$.getJSON(call, '', process_approved);

gurubobnz: Yeah inline

Submitted by Anonymous on Tue, 11/03/2009 - 07:53.

gurubobnz: Yeah inline function ftw when it comes to jQuery and JSON. Especially if this is the only plac you call your process_data() function.

The problem here is simple,

Submitted by Anonymous on Tue, 09/01/2009 - 15:21.

The problem here is simple, but maybe thats because I've spent so much time in javascript land...
you've wrapped your function reference in string literals. So its no longer a reference to your function its a string.

also the second argument is optional (jQuery must do some type checking on its arguments) so you can go back to the two-liner :D


var call = '..';
$.getJSON(call, process_approved);

Hope this helps.

string != function

Submitted by Anonymous on Sat, 07/25/2009 - 21:05.

Assuming you've been cut'n'pasting accurately from your source code, the Javascript engine was entirely correct - you passed a string containing the name of the function:

$.getJSON(call, '', 'process_approved');

instead of a reference to the function itself:

$.getJSON(call, '', process_approved);

There should be no logical difference between using that vs. using an anonymous function like you did in your workaround... except the workaround is larger and marginally slower.

[end of drive-by-debugging session]

KTHXBYE

getJSON

Submitted by Anonymous on Mon, 06/22/2009 - 15:46.

You might also wish to try

$.getJSON(call,'',process_approved);

ie: without the '' marks.

Assuming that function exists at the declaration/calling of that line of code it should work as prescribed.

Using string references to function names ( that is, with ' marks ) is generally not recommended.

Functions in JavaScript are "First Class" objects, which mean they can be literally passed around in variables.