我有我要显示的字符串变量final String wrongPw = "Wrong Password";
我的AJAX是按钮:
AjaxButton yesButton = new AjaxButton("yesButton", yesNoForm) {
private static final long serialVersionUID = -3827487963204274386L;
@Override
protected void onSubmit(AjaxRequestTarget target, Form form) {
if (target != null && password.equals(getPw())) {
answer.setAnswer(true);
modalWindow.close(target);
}else if(target != null && !password.equals(getPw())){
answer.setAnswer(false);
wrongPW.setVisible(true);
}
}
};
再向下:
wrongPW.setVisible(false);
add(wrongPW);
当我点击是按钮时,我必须刷新页面以显示
wrongPW
。如何动态地做到这一点?
最佳答案
您必须像这样将要更新的组件添加到目标:
target.add(wrongPW);
确保在要动态更改其可见性的组件初始化时设置标记占位符标签,否则Wicket将找不到它。
wrongPW.setOutputMarkupPlaceholderTag(true);
解释原因如下:https://stackoverflow.com/a/9671796/2795423