I'm getting a bit confused how to redirect to a new page after a form submission using AjaxFormSubmitBehavior.

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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