Action +ajax 不跳转问题！

Question

public class LoginAction extends ActionSupport{
protected HttpServletRequest request=ServletActionContext.getRequest();
protected HttpServletResponse response=ServletActionContext.getResponse();
PrintWriter out;
@Override
public String execute() throws Exception {
String Do_user=(String)request.getParameter("Do_user");
System.out.println("传输测试"+Do_user);
//1登录 2注册
if(Do_user.equals("1")){
String user=(String)request.getParameter("user");
String password=(String)request.getParameter("password");
Boolean A=jdbc_url.Douser("user", "uname",user);
Boolean B=jdbc_url.Douser("user", "pswrod",password);
System.out.println("执行完成");
}
** return "table"**;
}

   <action name="login" class="controller1.LoginAction">
            <result name="blank">com/blank.jsp</result>
            <result name="chart">com/chart.jsp</result>
            <result name="form">com/form.jsp</result>
            <result name="main_index">com/main_index.jsp</result>
            <result name="registeration">com/registeration.jsp</result>
            <result name="ta-panel">com/tab-panel.jsp</result>
            <result name="table">com/table.jsp</result>
            <result name="ui">com/ui.jsp</result>
            <result name="index">login.jsp</result>

        </action>

怎么不跳转了？

Answer 1

你是用ajax的方式调用的LoginAction？那么你需要页面自己去跳转，比如
$ajax.post(地址, 数据,
success: function(data){
window.location.href=data;
});
这里，假设你返回的是跳转的地址