[jQuery] Re: value visible in ajax response, but not used in output

[dcrunche]

Hi Scott,

thanks, that was what did the trick. I didn't need this without using
the actual vars - then it worked well. But it makes sense I need to
actually return the new result.

D.

On Nov 25, 5:57 pm, Scott Sauyet  wrote:
> On Nov 25, 11:12 am, dcrunche  wrote:
>
> > When I check the response given in the console I see the output I
> > want, but that is not showing on the page.
>
> > Here's the JS I use:http://pastie.org/714683
>
> I don't think you are actually using the resulting data anywhere in
> that code, except in the outermost success function.  For each of the
> (three levels of!) nested .ajax calls, you ignore the returning data
> parameter.
>
> Perhaps you just need to copy this line to each of them:
>
>     $('#productWizard fieldset').html(data);
>
> Cheers,
>
>   -- Scott

[jQuery] value visible in ajax response, but not used in output

[dcrunche]

hi,

I am rather new to jquery & ajax but I managed to get some simple
stuff done.
Now I am working on a wizard form that uses select items (with the
options pulled from a database).

This works nice. Now I want to take it a step further: I want the
selection of the second selector be based on the value of the first
selector. The third selector must be based on the first and the
second.

When I check the response given in the console I see the output I
want, but that is not showing on the page.

I have no clue where to find that...

Here's the JS I use: http://pastie.org/714683
Here's the PHP file I use: http://pastie.org/714685

If someone could take a look, that would be very nice.
Thanks in advance.