封装AjaxResult，返回后台对象

1.AjaxResult一般使用 success/msg即可，但有时需求需要返回后台对象时，需要进行如下操作

/**
 * 返回结果
 */
public class AjaxResult {

 ???public static void main(String[] args) {
 ???????//封装AjaxResult--返回后台数据
 ???????AjaxResult.me().setSuccess(true).setMsg("恭喜你操作成功").setObject("你运气好，账号密码都是对的");
 ???}

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

 ???private Boolean success = true;//默认操作成功
 ???private String msg = "操作成功";//返回前端操作的文字结果

 ???private Object object;//返回后台的对象

 ???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.测试

public class UserController {
 ???/*
 ???????@RequestBody主要用来接收前端传递给后端的json字符串中的数据的
 ????*/
 ???@RequestMapping(value = "/login",method = RequestMethod.POST)
 ???public AjaxResult login(@RequestBody User user){
 ???????//模拟登录--假设已经从数据库中查出来账号密码
 ???????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("登录成功").setObject(null);
 ???????????}
 ???????}
 ???????return AjaxResult.me().setSuccess(false).setMsg("登录失败").setObject("账号密码不正确");
 ???}
}

封装AjaxResult，返回后台对象

原文地址：https://www.cnblogs.com/wgyi140724-/p/10604664.html