On Friday, November 16, 2012 1:53:13 AM UTC-5, Benjamin Such 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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