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action中无法获取表单传回的username值

action中无法获取表单传回的username值

落子鸢 2017-03-12 14:58:57
public class User {  private String userName;  private String password;  public String getuserName() {   return userName;  }  public void setuserName(String name) {   this.userName = name;  }  public String getPassword() {   return password;  }  public void setPassword(String password) {   this.password = password;  }   }public class UserAction { private User user;  public User getUser() {  return user; } public void setUser(User user) {  this.user = user; } public String login(){  ActionContext context = ActionContext.getContext();  Map<String, Object> session= context.getSession();    session.put("userName", user.getuserName());  System.out.println(user.getuserName());  System.out.println("登录");  return Action.SUCCESS; }}<struts>    <constant name="struts.enable.DynamicMethodInvocation" value="false" />       <constant name="struts.devMode" value="true" />    <package name="default" namespace="/" extends="struts-default">  <action name="user_*" class="jsh.struts.action.UserAction" method="{1}" >   <result name="success">    success.jsp   </result>   <result name="error">    fail.jsp   </result>  </action>  <action name="helloworld" class="jsh.struts.action.UserAction">   <result>    index.jsp   </result>  </action>    </package></struts>表单<form action="user_login" class="jsh.struts.action.UserAction">   用户名:<input  type="text" value="" name="user.userName"><br/>   密码:<input type="text" name="user.password"><br/>   <input type= "submit" value = "登录">  </form>下面的是控制台的信息:[http-nio-8080-exec-2] ERROR com.opensymphony.xwork2.interceptor.ParametersInterceptor - Developer Notification (set struts.devMode to false to disable this message):Unexpected Exception caught setting 'user.userName' on 'class jsh.struts.action.UserAction: Error setting expression 'user.userName' with value ['jia', ]null登录
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慕瓜9220888

TA贡献868条经验 获得超0个赞

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反对 回复 2021-10-19
?
梦影剑魂

TA贡献66条经验 获得超21个赞

可能是user没有实例化,你可以在action类中实现ModelDriven<T> 接口,然后把user实例化:private  User=new User();

再在getmodel方法中返回该User

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反对 回复 2017-03-12
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