Java 在 Spring Boot 中返回 JSON 对象作为响应

Returning JSON object as response in Spring Boot

提问by iwekesi

I have a sample RestController in Spring Boot:

我在 Spring Boot 中有一个示例 RestController：

@RestController
@RequestMapping("/api")
class MyRestController
{

    @GetMapping(path = "/hello")
    public JSONObject sayHello()
    {
        return new JSONObject("{'aa':'bb'}");
    }


}

I am using the JSON library org.json

我正在使用 JSON 库 org.json

When I hit API /hello, I get an exception saying :

当我点击 API 时/hello，我收到一个异常说：

Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is java.lang.IllegalArgumentException: No converter found for return value of type: class org.json.JSONObject] with root cause

java.lang.IllegalArgumentException: No converter found for return value of type: class org.json.JSONObject

servlet [dispatcherServlet] 的 Servlet.service() 在路径为 [] 的上下文中抛出异常 [请求处理失败；嵌套异常是 java.lang.IllegalArgumentException: No converter found for return value of type: class org.json.JSONObject] 与根本原因

java.lang.IllegalArgumentException：找不到类型为 org.json.JSONObject 的返回值的转换器

What is the issue? Can someone explain what exactly is happening?

问题是什么？有人能解释一下到底发生了什么吗？

采纳答案by prem kumar

As you are using Spring Boot web, Hymanson dependency is implicit and we do not have to define explicitly. You can check for Hymanson dependency in your pom.xmlin the dependency hierarchy tab if using eclipse.

当您使用 Spring Boot web 时，Hymanson 依赖是隐式的，我们不必明确定义。pom.xml如果使用 eclipse ，您可以在依赖层次结构选项卡中检查 Hymanson 依赖。

And as you have annotated with @RestControllerthere is no need to do explicit json conversion. Just return a POJO and Hymanson serializer will take care of converting to json. It is equivalent to using @ResponseBodywhen used with @Controller. Rather than placing @ResponseBodyon every controller method we place @RestControllerinstead of vanilla @Controllerand @ResponseBodyby default is applied on all resources in that controller.
Refer this link:https://docs.spring.io/spring/docs/current/spring-framework-reference/html/mvc.html#mvc-ann-responsebody

正如您所注释的那样，@RestController无需进行显式的 json 转换。只需返回一个 POJO，Hymanson 序列化程序将负责转换为 json。@ResponseBody与@Controller一起使用时相当于使用。我们不是放置@ResponseBody在每个控制器方法上，而是放置在@RestController原版上，@Controller并且@ResponseBody默认情况下应用于该控制器中的所有资源。
请参阅此链接：https : //docs.spring.io/spring/docs/current/spring-framework-reference/html/mvc.html#mvc-ann-responsebody

The problem you are facing is because the returned object(JSONObject) does not have getter for certain properties. And your intention is not to serialize this JSONObject but instead to serialize a POJO. So just return the POJO.
Refer this link:https://stackoverflow.com/a/35822500/5039001

您面临的问题是因为返回的对象（JSONObject）没有某些属性的 getter。您的意图不是序列化这个 JSONObject，而是序列化一个 POJO。所以只需返回POJO。
请参阅此链接：https : //stackoverflow.com/a/35822500/5039001

If you want to return a json serialized string then just return the string. Spring will use StringHttpMessageConverter instead of JSON converter in this case.

如果你想返回一个 json 序列化的字符串，那么只需返回该字符串。在这种情况下，Spring 将使用 StringHttpMessageConverter 而不是 JSON 转换器。

回答by g00glen00b

The reason why your current approach doesn't work is because Hymanson is used by default to serialize and to deserialize objects. However, it doesn't know how to serialize the JSONObject. If you want to create a dynamic JSON structure, you can use a Map, for example:

您当前的方法不起作用的原因是因为默认情况下使用 Hymanson 来序列化和反序列化对象。但是，它不知道如何序列化JSONObject. 如果要创建动态 JSON 结构，可以使用Map，例如：

@GetMapping
public Map<String, String> sayHello() {
    HashMap<String, String> map = new HashMap<>();
    map.put("key", "value");
    map.put("foo", "bar");
    map.put("aa", "bb");
    return map;
}

This will lead to the following JSON response:

这将导致以下 JSON 响应：

{ "key": "value", "foo": "bar", "aa": "bb" }

This is a bit limited, since it may become a bit more difficult to add child objects. Hymanson has its own mechanism though, using ObjectNodeand ArrayNode. To use it, you have to autowire ObjectMapperin your service/controller. Then you can use:

这有点受限制，因为添加子对象可能会变得更加困难。Hymanson 有自己的机制，使用ObjectNode和ArrayNode。要使用它，您必须ObjectMapper在您的服务/控制器中自动装配。然后你可以使用：

@Autowired
ObjectMapper mapper;
(...)
@GetMapping
public ObjectNode sayHello() {
    ObjectNode objectNode = mapper.createObjectNode();
    objectNode.put("key", "value");
    objectNode.put("foo", "bar");
    objectNode.put("number", 42);
    return objectNode;
}

This approach allows you to add child objects, arrays, and use all various types.

这种方法允许您添加子对象、数组并使用所有各种类型。

回答by Sangam Belose

You can either return a response as Stringas suggested by @vagaasen or you can use ResponseEntityObject provided by Spring as below. By this way you can also return Http status codewhich is more helpful in webservice call.

您可以String按照@vagaasen 的建议返回响应，也可以使用ResponseEntitySpring 提供的对象，如下所示。通过这种方式，您还可以返回Http status code这对 web 服务调用更有帮助。

@RestController
@RequestMapping("/api")
public class MyRestController
{

    @GetMapping(path = "/hello", produces=MediaType.APPLICATION_JSON_VALUE)
    public ResponseEntity<Object> sayHello()
    {
         //Get data from service layer into entityList.

        List<JSONObject> entities = new ArrayList<JSONObject>();
        for (Entity n : entityList) {
            JSONObject entity = new JSONObject();
            entity.put("aa", "bb");
            entities.add(entity);
        }
        return new ResponseEntity<Object>(entities, HttpStatus.OK);
    }
}

回答by Dmitry

@RequestMapping("/api/status")
public Map doSomething()
{
    return Collections.singletonMap("status", myService.doSomething());
}

PS. Works only for 1 value

附注。仅适用于 1 个值

回答by maximus

you can also use a hashmap for this

您也可以为此使用哈希图

@GetMapping
public HashMap<String, Object> get() {
    HashMap<String, Object> map = new HashMap<>();
    map.put("key1", "value1");
    map.put("results", somePOJO);
    return map;
}

回答by Ved Prakash

use ResponseEntity<ResponseBean>

用 ResponseEntity<ResponseBean>

Here you can use ResponseBean or Any java bean as you like to return your api response and it is the best practice. I have used Enum for response. it will return status code and status message of API.

在这里，您可以根据需要使用 ResponseBean 或 Any java bean 来返回 api 响应，这是最佳实践。我使用 Enum 进行响应。它将返回 API 的状态代码和状态消息。

@GetMapping(path = "/login")
public ResponseEntity<ServiceStatus> restApiExample(HttpServletRequest request,
            HttpServletResponse response) {
        String username = request.getParameter("username");
        String password = request.getParameter("password");

        loginService.login(username, password, request);
        return new ResponseEntity<ServiceStatus>(ServiceStatus.LOGIN_SUCCESS,
                HttpStatus.ACCEPTED);
    }

for response ServiceStatus or(ResponseBody)

用于响应 ServiceStatus 或（ResponseBody）

    public enum ServiceStatus {

    LOGIN_SUCCESS(0, "Login success"),

    private final int id;
    private final String message;

    //Enum constructor
    ServiceStatus(int id, String message) {
        this.id = id;
        this.message = message;
    }

    public int getId() {
        return id;
    }

    public String getMessage() {
        return message;
    }
}

Spring REST API should have below key in response

Spring REST API 应该具有以下键作为响应

1. Status Code
2. Message

1. 状态码
2. 信息

you will get final response below

您将在下面得到最终答复

{

   "StatusCode" : "0",

   "Message":"Login success"

}

you can use ResponseBody(java POJO, ENUM,etc..) as per your requirement.

您可以根据您的要求使用 ResponseBody(java POJO、ENUM 等)。