Re: Bad Request "URL not found" - after an JQuery change event

Cake throws a 404 when debugging is disabled. You probably have an
error somewhere in your PHP, not the JS.

On Fri, Nov 16, 2012 at 1:53 AM, Benjamin Such <benjamin.such@gmail.com> wrote:
> Hey guys, I am really stuck with that problem. I have a JQuery Dialog which
> contains a selectbox. When I change the option inside this selectbox,
> another selectbox shows up with options generated from an .ajax() event.
> Anyway... everything works fine, but when I submit the form CakePHP says:
> 404 Bad Request - URL was not found, but it DOES EXISTS 100%. I tested it
> several times to be sure, but another thing is: When submit the form WITHOUT
> selecting any option from the first selectbox the form is submitted
> correctly. I dont know what to post really but here is my Javascript:
>
>> $('select[data-onchange=true]').live("change", function() {
>>
>> var _this = $(this);
>> var element = $('select[data-onchange-trigger=' + _this.attr("id")
>> + ']:first');
>>
>> if (typeof element == "object") {
>> if (_this.attr("data-onchange-load")) {
>> var options = element.prop("options");
>> var load = _this.attr("data-onchange-load");
>> $.ajax({
>> dataType: "json",
>> url: load + _this.find('option:selected').val(),
>> beforeSend: function() {
>> $('.ajax-loader').show();
>> element.find('option').remove();
>> element.parent().hide();
>> },
>> success: function(data) {
>> $.each(data, function(index, _data) {
>> if (typeof _data == "object") {
>> $.each(_data, function(index, _data) {
>> options[options.length] = new
>> Option(_data.name, _data.id);
>> });
>> }
>> });
>> element.parent().show();
>> },
>> complete: function() {
>> $('.ajax-loader').hide();
>> }
>> });
>> }
>> }
>> });
>
>
> The error occurs only when I submit the form after I change the option on
> the selectbox.
>
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