封裝AjaxResult，返回后台對象

1.AjaxResult一般使用 success/msg即可，但有時需求需要返回后台對象時，需要進行如下操作

/**
 * 返回結果
 */
public class AjaxResult {

 public static void main(String[] args) {
 //封裝AjaxResult--返回后台數據
 AjaxResult.me().setSuccess(true).setMsg("恭喜你操作成功").setObject("你運氣好，賬號密碼都是對的");
 }

 public static AjaxResult me(){
 return new AjaxResult();
 }

 private Boolean success = true;//默認操作成功
 private String msg = "操作成功";//返回前端操作的文字結果

 private Object object;//返回后台的對象

 public Boolean getSuccess() {
 return success;
 }

 public AjaxResult setSuccess(Boolean success) {
 this.success = success;
 return this;
 }

 public String getMsg() {
 return msg;
 }

 public AjaxResult setMsg(String msg) {
 this.msg = msg;
 return this;
 }

 public Object getObject() {
 return object;
 }

 public AjaxResult setObject(Object object) {
 this.object = object;
 return this;
 }

 @Override
 public String toString() {
 return "AjaxResult{" +
 "success=" + success +
 ", msg='" + msg + '\'' +
 ", object=" + object +
 '}';
 }
}

2.測試

public class UserController {
    /*
 @RequestBody主要用來接收前端傳遞給后端的json字符串中的數據的
 */
 @RequestMapping(value = "/login",method = RequestMethod.POST)
 public AjaxResult login(@RequestBody User user){
 //模擬登錄--假設已經從數據庫中查出來賬號密碼
 if(user!=null&& !StringUtils.isEmpty(user.getName())&&!StringUtils.isEmpty(user.getPassword())){
 if("admin".equals(user.getName())&&"1234".equals(user.getPassword())){
 return AjaxResult.me().setSuccess(true).setMsg("登錄成功").setObject(null);
 }
 }
 return AjaxResult.me().setSuccess(false).setMsg("登錄失敗").setObject("賬號密碼不正確");
 }
}