Well the error messages kind of gives it away, doesn't it?
But to explain further: setDefaultFormProcessing(false) doesn't mean the
button can be placed outside a a form. It means that the default
validation and submit behavior of the form is not executed when pressing
the button. Why do you want to use AjaxButton? Outside a form AjaxLink
does the job as you already discovered...
On Thu, Mar 28, 2013, at 18:45, eugenebalt wrote:
> I have the below block of code. A Link or an AjaxLink works perfectly
> here
> inside a WebMarkupContainer. But when I use an AjaxButton (with a
> setDefaultFormProcessing=False), I get an error. Thanks
>
> AjaxButton linkCopy = new AjaxButton("linkCopy") {
>
> @Override
> protected void onSubmit(AjaxRequestTarget arg0, Form<?>
> arg1) {
> // TODO Auto-generated method stub
>
> }
> };
> linkCopy.setDefaultFormProcessing(false);
> wmcMarkup.add(linkCopy);
>
> Error:
> java.lang.IllegalStateException: form was not specified in the
> constructor
> and cannot be found in the hierarchy of the component this behavior is
> attached to
> at
> org.apache.wicket.ajax.form.AjaxFormSubmitBehavior.getForm(AjaxFormSubmitBehavior.java:92)
> at
> org.apache.wicket.ajax.form.AjaxFormSubmitBehavior.getEventHandler(AjaxFormSubmitBehavior.java:108)
> at
> org.apache.wicket.ajax.markup.html.form.AjaxButton$1.getEventHandler(AjaxButton.java:131)
> at
> org.apache.wicket.ajax.AjaxEventBehavior.onComponentTag(AjaxEventBehavior.java:111)
> at
> org.apache.wicket.behavior.AbstractAjaxBehavior.onComponentTag(AbstractAjaxBehavior.java:134)
> at org.apache.wicket.Component.renderComponentTag(Component.java:4009)
> at org.apache.wicket.Component.renderComponent(Component.java:2610)
>
>
>
>
> --
> View this message in context:
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> Sent from the Users forum mailing list archive at Nabble.com.
>
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