you can use getinput() to get the raw value or override wantOnSelectionChangedNotifications() to return true and implement a listener in onSelectionChanged, but then its not ajax.
-igor On Thu, Oct 15, 2009 at 6:02 AM, Christian Reiter <c.rei...@gmx.net> wrote: > Hello! > > I've got a DropDownChoice in one of my forms, with which the user can select > a icon. > I want to display a preview of this icon beneath the drop down choice box. > Therefore I added an AjaxFormValidatingBehavior to the DropDownChoice. Now, > when > the user changes the value of the DropDownChoice I get informed in the > onError or onSubmit > method of the AjaxFormValidatingBehavior. Everything works fine if the form > validation is > ok and the form data gets synced to my data model (then I query my model for > the selected icon > and update the corresponding Image and send it to the client). > > The problem occurs when the form validation doesn't work, because in this > case, the > form data isn't synced to the model and I have no chance to know which icon > was selected > by the user (through the DropDownChoice). As I want to display a preview of > the icon also > if any of the previous form components didn't validate successful, I am in > need of a method > which tells me which Icon was chosen by the user, regardless of the previous > validation > result. > > Is this possible? > > Thanks in anticipation! > > Best Regards, > > chris > > > -- > > > Christian Reiter |||c.rei...@gmx.net > > --------------------------------------------------------------------- > To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org > For additional commands, e-mail: users-h...@wicket.apache.org > > --------------------------------------------------------------------- To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
