struts2中的execute（）方法中的好玩问题，不知道为什么，求解答。

用的ssh框架完全没有问题，主要问题是execute（）方法中if语句问题。直接上代码奔主题去啦。

 package cn.web.action;

import com.opensymphony.xwork2.ActionSupport;

import cn.web.daoimp.UserDaoImp;
import cn.web.service.Users;

public class UserAction extends ActionSupport {
  private  String username;
  private  String password;
  public boolean isa;

  public boolean isa(){
      isa="a".equals(username);
      return isa;
  }


  public boolean isIsa() {
    return isa;
}

public void setIsa(boolean isa) {
    this.isa = isa;
}

public UserAction(){}

  public UserAction(String username,String password){

      this.username=username;
      this.password=password;
  }



  public String getUsername() {
        return username;
    }
    public void setUsername(String username) {
        this.username = username;
    }
    public String getPassword() {
        return password;
    }
    public void setPassword(String password) {
        this.password = password;
    }


    public String execute() {

        if (("a".equals(username))){
             return "success";
            }
           else {
               return "error";
           }

    }

这是action的主要代码，就是想实现登陆功能什么的。但是上面代码的execute（）方法中的if判断语句有问题。直接提取出来。

 public String execute() {

        if (("a".equals(username))){
             return "success";
            }
           else {
               return "error";
           }

    }

无论我在前端输入a还是其他的都会返回“error”。但是当我把if（）语句改成这样

 public boolean isa;

  public boolean isa(){
      isa="a".equals(username);
      return isa;
  }

 public String execute() {

        if (isa()){
             return "success";
            }
           else {
               return "error";
           }

    }

改写之后我就能得到正确的我想要的结果。我不明白这是为什么，求大神指点。

package Action;

import com.opensymphony.xwork2.ActionSupport;

/**

• Created by Administrator on 2016/11/29.
  */
  public class netTestAction extends ActionSupport{
  private String username;
  private String password;

  public String getUsername() {
  return username;
  }

  public void setUsername(String username) {
  this.username = username;
  }

  public String getPassword() {
  return password;
  }

  public void setPassword(String password) {
  this.password = password;
  }

  @Override
  public String execute() throws Exception {

  System.out.println(username + "::::" +password);
  if (("a".equals(username))){
      return "success";
  }
  else {
      return "error";
  }
  

  }

}

写个sout看一眼那个username 是啥嘛，万一不是“a”是别的什么呢