[support] jQuery.ajax returns 404

[Seth Freach]

Scott,

Using firebug in Firefox will let you see exactly what goes back and 
forth in an ajax request.

However, below will not work because, as a security feature, ajax 
requests can only be sent to the same domain that the page originated 
from.  Ie, if your site is example.com, then ajax GETs and PUTs can only 
be performed back to example.com, not to other sites directly from the 
client browser such as api2.citysearch.com.

What you need to do is make a server side wrapper.  Your ajax call 
should be back to your own site, example.com, which can then take the 
request and with PHP build a curl request to api2.citysearch.com.  Once 
your server has received the response from citysearch, then it can 
deliver it as the reply to the original ajax request.

Seth

Scott wrote:
> I'm still experimenting with a new citysearch module.  I've managed to
> fetch back the json with curl but I'd still like to let ajax handle it.
> I have this function but it reports a 404 error.  Is there a way to
> determine the url value actually being sent?  The generated value I'm
> using here works both with window open() and with my curl function.
> My code:
>
> function processSearch(form) {
>   var what = js_trim(form.what.value);
>   var where = js_trim(form.where.value);
>   var loc = js_trim(form.location_or_event.value);
>   var publisher = js_trim(form.publisher.value);
>   var api_key = js_trim(form.api_key.value);
>
>   var URL =  'api2.citysearch.com/search/'+loc+'?what='+what
> +'&where='+where+'&format=json&publisher='+publisher
> +'&api_key='+api_key;
>
> alert(URL);
> $.ajax({
>   type: 'get',
>   url: URL,
>   dataType: 'jsonp',
>   success: function() { alert("Some success"); },
>   error:function (xhr, ajaxOptions, thrownError){
>                     alert(xhr.status);
>                     alert(xhr.statusText);
>                     alert(xhr.responseText);
>                 }
> }); // end ajax method
> }
>
> function js_trim(str) {
>   return str.replace(/^\s\s*/, '').replace(/\s\s*$/, '');
> }
>
>
>