Struts里ActionErrors saveErrors()问题，谁能帮我？

[rainbow616]

package hello;

import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpSession;
import javax.servlet.http.HttpServletResponse;

import org.apache.struts.action.Action;
import org.apache.struts.action.ActionMessage;
import org.apache.struts.action.ActionMessages;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
import org.apache.struts.util.MessageResources;

public final class HelloAction extends Action {

public ActionForward execute(ActionMapping mapping,
ActionForm form,
HttpServletRequest request,
HttpServletResponse response)
throws Exception {
MessageResources messages = getResources(request);

ActionMessages errors = new ActionMessages();
String userName = ((HelloForm) form).getUserName();

String badUserName = "Monster";

if (userName.equalsIgnoreCase(badUserName)) {
errors.add("username", new ActionMessage("hello.dont.talk.to.monster", badUserName ));
saveErrors(request, errors);这句话出错了，我要怎么解决？
return (new ActionForward(mapping.getInput()));
}


PersonBean pb = new PersonBean();
pb.setUserName(userName);
pb.saveToPersistentStore();


request.setAttribute( Constants.PERSON_KEY, pb);

// Remove the Form Bean - don't need to carry values forward
request.removeAttribute(mapping.getAttribute());

// Forward control to the specified success URI
return (mapping.findForward("SayHello"));

}
}

[fuliqiang]

不是吧，还没明白？

ActionMessages errors = new ActionMessages();
这句话你声明了一个ActionMessages对象。

然后在这个方法里传入这个对象。
saveErrors(request, errors);
但是saveErrors()要求接收的是ActionErrors类的对象。
而不是ActionMessages类的对象
我估计这个例子是错的。

你把ActionMessages errors = new ActionMessages();
改成
ActionErrors errors=new ActionErrors();
试试