java spring项目controller中接收一个string，和一个List为入参报错,前端参数应该怎么传？

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。