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我想刷新 AjaxButton 的 onSubmit(AjaxRequestTarget target, Form form) 上的整个表单。这怎么可能做任何评论。
在表单构造函数中,或者在添加表单之前,将表单设置为输出标记 id:
form.setOutputMarkupId(true);
然后onSubmit将您的表单添加到target.
onSubmit
target
target.add(form);