Hi, Do you have FeedbackPanel in the page ? And is it Ajax updated when there is an error ? It will show you the error. You need to debug #onError(AjaxRequestTarget, Form) instead of #onError().
Martin Grigorov Wicket Training and Consulting On Thu, Mar 20, 2014 at 1:40 PM, chathuraka.waas <[email protected]>wrote: > hi, > > i'm trying to migrate my applicatioin to 6.14, > > i'm having a form which has a AjaxSubmitLink. i have attached a form > validator to the form as well. for debugging purposes i have placed debug > points at validator method,inside onSubmit() of AjaxSubmit link and inside > onError() of AjaxSubmitLink. > > When i have the validator added to the form and debug i reach to the > validator method but doesnt get into onsubmit or onError methods. > > when i remove the validator from the form the flow reaches onError method > on > the submitAjaxLink. but i'm unable to find whats causing the error. > > is there a way i can identify the error being caused so that i can rectify > it. There are no expceptions thrown or compilation errors. > > Thanks in advance. > > > > > -- > View this message in context: > http://apache-wicket.1842946.n4.nabble.com/Cannot-find-the-form-error-caused-tp4665039.html > Sent from the Users forum mailing list archive at Nabble.com. > > --------------------------------------------------------------------- > To unsubscribe, e-mail: [email protected] > For additional commands, e-mail: [email protected] > >
