Java spring boot -返回json响应

Question

我想为一个新的Java Spring Boot应用程序创建一个自定义的json响应。现在，我只想创建一个虚拟的json对象--开始控制输出json响应的结构。

当我这样做时，它想要创建.puts的cast前缀--我曾经使用Mongodb -- BasicDBObject --但是现在我这里没有mongo。

 DBObject obj = new BasicDBObject();
 obj.put( "foo", "bar" );

我该怎么做呢？

--

@RequestMapping(value = "/login", method = RequestMethod.GET)
@CrossOrigin(origins = {"*"})
public ResponseEntity<?> login(
        @RequestParam(value="email", required=false, defaultValue="email") String email,
        @RequestParam(value="password", required=false, defaultValue="password") String password, 
        HttpServletRequest request
        ) throws  Exception {

                Object response = new Object();
                    response.put("information", "test");
                    response.put("id", 3);
                    response.put("name", "course1");

                return new ResponseEntity<>(response, HttpStatus.OK);
        }

Answer 1

**@ResetController**   // add this
public class YourClass{

    @RequestMapping(value = "/login", method = RequestMethod.GET)
    @CrossOrigin(origins = {"*"})
    public ResponseEntity<?> login(){
        Object response = new Object();
                    response.put("information", "test");
                    response.put("id", 3);
                    response.put("name", "course1");

        return new ResponseEntity<>(response, HttpStatus.OK);
     }
}