I use the onSubmit method of AjaxFormSubmitBehavior component to notify user for special message, for example:
buttonSave.add( new AjaxFormSubmitBehavior ( form, "onclick" ) {
protected void onSubmit(AjaxRequestTarget target) {
if( !getFeedbackMessages().isEmpty() ) {
target.addJavascript( "alert('Validation failed')" );
}
}
});
and the form onSubmit to save data and redirect to the next page:
form = new Form("form1") {
void onSubmit() {
if( findSubmittingButton() == buttonSave ) {
dao.save( object );
setResponsePage( NextPage.class );
}
}
};
The problem is that in this way I will get the response page as response content of the Ajax call!
How to redirect to another page using AjaxFormSubmitBehavior?
Thanks for helping.
- Paolo
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