java spring项目controller中接收一个string，和一个List为入参报错,前端

Question

只接收一个List作为入参的话不报错，下方为代码：前台：$.ajax({    type:"post",    url: 'add',    contentType:"application/json; charset=utf-8",    dataType:"json",    data:JSON.stringify([{id:1,name:"hello"},{id:2,name:"hello"}]),    success:function(data){        console.log(data);    }})后台：@ResponseBody@RequestMapping(value = "add", consumes = "application/json; charset=utf-8")public String add(@RequestBody List<Tag> param) {    System.out.println("param:" + param);    return "成功";}前台控制台显示的数据格式为：[{id: 1, name: "hello"}, {id: 2, name: "hello"}]0:{id: 1, name: "hello"}1:{id: 2, name: "hello"}那如果我后台想接受一个字符串和一个list该怎么写呢，后台代码改为：@ResponseBody@RequestMapping(value = "add", consumes = "application/json; charset=utf-8")public String add(@RequestBody String content, @RequestBody List<Tag> param) {前台应该怎么写？

Answer 1

两种解决方案

1、

如果参数比较少 直接url?parm=value,

那么就用@RequestParam注解

2、封装成对象

    class A {

        private String parm1;

        private String parm2;

        private List<B> list;

    }

对应的前台

var data =new Object;

data.param1="value";

data.param2="value";

var  list =  new Array();

....省略list的数据初始化

data.list = list;

ajax的data

 data:JSON.stringify(data)

Answer 2

@ResponseBody 接收的是前台你用 JSON 传递过来的一个整体，所以你前台需要将 String 和 List 封装在一个对象里面。

var json = {};

json['content'] = content;

json['param'] = [{id: 1, name: "hello"}, {id: 2, name: "hello"}];

// ajax data

data:JSON.stringify(json)

后台接收的话，建议是写一个对象来接收比如，

public class SomeModel {

    private String content;

    private List<Tag> param;

    // getter and setter

}

controller

@RequestMapping(value = "add", consumes = "application/json; charset=utf-8")

public String add(@RequestBody SomeModel someModel) {

    String content = someModel.getContent();

    List<Tag> param = someModel.getParam();

}

或者可以偷懒，用 Map 来接收。