ok, seems like i got it on my own, with a bunch of help from malsup's
blog and the jq docs.
looks like that. that's ok, you think?
(function($) {
jQuery.fn.jsonForm = function(options) {
var opts = $.extend({}, $.fn.jsonForm.defaults, options);
return this.each(function(){
jQuery(this).ajaxForm({
type: "json",
success: function(data){
// private implementation here ...
opts.success(data);
}
});
});
};
$.fn.jsonForm.defaults = {
success: function(){}
};
})(jQuery);
On Feb 9, 5:20 pm, Jan Limpens <jan.limp...@gmail.com> wrote:
> When I try to do this, firebug does not alert me of anything wrong,
> the form is posted normally (no ajax).
> What mistake am I making?
>
> // requires jquery.forms plugin
> jQuery.fn.jsonForm = function(onSuccess) {
> this.ajaxForm({
> dataType: 'json',
> success: function(data) {
> if (data.Message) {
> alert(data.Message); // my json reply always has these 2
> fields
> }
> if (data.ActionUrl != null) {
> top.location = data.ActionUrl;
> }
> if (onSuccess != null) {
> onSuccess(data); // what if onSuccess has an empty
> signature?
> }
> }
>
> });
> }
>
> So I can call it like so:
>
> var onSuccess = function(data){
> $.each(data.Customer, function(i, item){
> $('#searchResultCustomers').append("<dd>"+ item.Name
> +"</dd>");
> });
> });
> $('.jsonFormCustomer').jsonForm(onSuccess);
>
> How could I call jsonForm without a null param, too?
>
> Help greatly appreciated!
