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&nbsp; &nbsp; class A {&nbsp; &nbsp; &nbsp; &nbsp; private String parm1;&nbsp; &nbsp; &nbsp; &nbsp; private String parm2;&nbsp; &nbsp; &nbsp; &nbsp; private List<B> list;&nbsp; &nbsp; }对应的前台var data =new Object;data.param1="value";data.param2="value";var&nbsp; list =&nbsp; new Array();....省略list的数据初始化data.list = list;ajax的data&nbsp;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 datadata: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 {&nbsp; &nbsp; private String content;&nbsp; &nbsp; private List<Tag> param;&nbsp; &nbsp; // getter and setter}controller@RequestMapping(value = "add", consumes = "application/json; charset=utf-8")public String add(@RequestBody SomeModel someModel) {&nbsp; &nbsp; String content = someModel.getContent();&nbsp; &nbsp; List<Tag> param = someModel.getParam();}或者可以偷懒，用 Map 来接收。