当Wicket AjaxFormValidatingBehavior验证失败时阻止浏览器表单提交

重写 onError() 方法在窗体或 AjaxFormValidatingBehavior . 如果你这样做的行为,我不确定是否会阻止表单提交。

new AjaxFormValidatingBehavior( form, "onsubmit") {
   public void onSubmit() {}
   public void onError() {}
}

可能有点晚了,但答案是:

public class SomePage extends WebPage {

    private FeedbackPanel feedbackMessageError = new FeedbackPanel("feedbackTabAddEmpMesError", new ExactLevelFeedbackMessageFilter(FeedbackMessage.ERROR));

    public SomePage(String id) {

      final Form<Void> form = new Form<>("tabFormAddEmp");
      add(form);

      //Name textfield cannot be empty
      final FormComponent<String> tabAddEmpName = new RequiredTextField<>("tabAddEmpName", Model.of(""));
      tabAddEmpName.setLabel(Model.of("Name"));
      tabAddEmpName.setOutputMarkupId(true);

      //Salarynumber has to be minimal 10 char long
      final FormComponent<String> tabAddEmpLoon = new RequiredTextField<>("tabAddEmpLoon", Model.of(""));
      tabAddEmpLoon.add(new StringValidator(10, null)).setLabel(Model.of("Salarynumber"));
      tabAddEmpLoon.setOutputMarkupId(true);

      final Button button = new Button("tabFormAddEmpBut");

      form.add(tabAddEmpName , tabAddEmpLoon, button);

      button.add(new AjaxFormValidatingBehavior(form, "onclick") {

        @Override
        public void onError(AjaxRequestTarget target) {
             //Add feedbackpanel to your html and voila!
             target.add(feedbackMessageError);
        }

        @Override
        protected void onSubmit(AjaxRequestTarget target) { 
          //Do some logic over here
        }

      }

   }
}