437

我将 Spring MVC 用于一个简单的 JSON API,其@ResponseBody基础方法如下。(我已经有一个直接生成 JSON 的服务层。)

@RequestMapping(value = "/matches/{matchId}", produces = "application/json")
@ResponseBody
public String match(@PathVariable String matchId) {
    String json = matchService.getMatchJson(matchId);
    if (json == null) {
        // TODO: how to respond with e.g. 400 "bad request"?
    }
    return json;
}

问题是,在给定的场景中,响应 HTTP 400 错误的最简单、最干净的方法是什么?

我确实遇到过以下方法:

return new ResponseEntity(HttpStatus.BAD_REQUEST);

...但我不能在这里使用它,因为我的方法的返回类型是字符串,而不是 ResponseEntity。

4

14 回答 14

693

将您的返回类型更改为ResponseEntity<>,然后您可以在下面使用 400

return new ResponseEntity<>(HttpStatus.BAD_REQUEST);

并为正确的请求

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

更新 1

在 spring 4.1 之后,ResponseEntity 中的辅助方法可以用作

return ResponseEntity.status(HttpStatus.BAD_REQUEST).body(null);

return ResponseEntity.ok(json);
于 2013-04-27T10:01:23.930 回答
115

像这样的东西应该可以工作,我不确定是否有更简单的方法:

@RequestMapping(value = "/matches/{matchId}", produces = "application/json")
@ResponseBody
public String match(@PathVariable String matchId, @RequestBody String body,
            HttpServletRequest request, HttpServletResponse response) {
    String json = matchService.getMatchJson(matchId);
    if (json == null) {
        response.setStatus( HttpServletResponse.SC_BAD_REQUEST  );
    }
    return json;
}
于 2013-04-26T09:35:33.123 回答
57

不一定是最紧凑的方式,但非常干净的 IMO

if(json == null) {
    throw new BadThingException();
}
...

@ExceptionHandler(BadThingException.class)
@ResponseStatus(value = HttpStatus.BAD_REQUEST)
public @ResponseBody MyError handleException(BadThingException e) {
    return new MyError("That doesnt work");
}

编辑如果使用 Spring 3.1+,您可以在异常处理程序方法中使用 @ResponseBody,否则使用 aModelAndView或其他东西。

https://jira.springsource.org/browse/SPR-6902

于 2013-04-26T09:19:36.903 回答
49

我会稍微改变实现:

首先,我创建一个UnknownMatchException

@ResponseStatus(HttpStatus.NOT_FOUND)
public class UnknownMatchException extends RuntimeException {
    public UnknownMatchException(String matchId) {
        super("Unknown match: " + matchId);
    }
}

请注意@ResponseStatus的使用,它将被 Spring 识别ResponseStatusExceptionResolver。如果抛出异常,它将创建一个具有相应响应状态的响应。(我还冒昧地将状态代码更改为404 - Not Found我认为更适合此用例的状态代码,但HttpStatus.BAD_REQUEST如果您愿意,可以坚持使用。)


接下来,我将更MatchService改为具有以下签名:

interface MatchService {
    public Match findMatch(String matchId);
}

最后,我将更新控制器并委托给 SpringMappingJackson2HttpMessageConverter以自动处理 JSON 序列化(如果您将 Jackson 添加到类路径并添加@EnableWebMvc或添加<mvc:annotation-driven />到您的配置,则默认添加它,请参阅参考文档):

@RequestMapping(value = "/matches/{matchId}", produces = MediaType.APPLICATION_JSON_VALUE)
@ResponseBody
public Match match(@PathVariable String matchId) {
    // throws an UnknownMatchException if the matchId is not known 
    return matchService.findMatch(matchId);
}

请注意,将域对象与视图对象或 DTO 对象分开是很常见的。这可以通过添加一个返回可序列化 JSON 对象的小型 DTO 工厂轻松实现:

@RequestMapping(value = "/matches/{matchId}", produces = MediaType.APPLICATION_JSON_VALUE)
@ResponseBody
public MatchDTO match(@PathVariable String matchId) {
    Match match = matchService.findMatch(matchId);
    return MatchDtoFactory.createDTO(match);
}
于 2013-04-27T10:10:36.310 回答
37

这是一种不同的方法。创建一个Exception带有 注释的自定义@ResponseStatus,如下所示。

@ResponseStatus(code = HttpStatus.NOT_FOUND, reason = "Not Found")
public class NotFoundException extends Exception {

    public NotFoundException() {
    }
}

并在需要时扔掉。

@RequestMapping(value = "/matches/{matchId}", produces = "application/json")
@ResponseBody
public String match(@PathVariable String matchId) {
    String json = matchService.getMatchJson(matchId);
    if (json == null) {
        throw new NotFoundException();
    }
    return json;
}

在此处查看 Spring 文档:http: //docs.spring.io/spring/docs/current/spring-framework-reference/htmlsingle/#mvc-ann-annotated-exceptions

于 2016-08-05T16:20:39.640 回答
26

最简单的方法是扔一个ResponseStatusException

    @RequestMapping(value = "/matches/{matchId}", produces = "application/json")
    @ResponseBody
    public String match(@PathVariable String matchId, @RequestBody String body) {
        String json = matchService.getMatchJson(matchId);
        if (json == null) {
            throw new ResponseStatusException(HttpStatus.NOT_FOUND);
        }
        return json;
    }
于 2020-01-15T12:58:05.100 回答
24

正如一些答案中提到的,可以为要返回的每个 HTTP 状态创建一个异常类。我不喜欢必须为每个项目的每个状态创建一个类的想法。这是我想出的。

  • 创建一个接受 HTTP 状态的通用异常
  • 创建 Controller Advice 异常处理程序

让我们来看看代码

package com.javaninja.cam.exception;

import org.springframework.http.HttpStatus;


/**
 * The exception used to return a status and a message to the calling system.
 * @author norrisshelton
 */
@SuppressWarnings("ClassWithoutNoArgConstructor")
public class ResourceException extends RuntimeException {

    private HttpStatus httpStatus = HttpStatus.INTERNAL_SERVER_ERROR;

    /**
     * Gets the HTTP status code to be returned to the calling system.
     * @return http status code.  Defaults to HttpStatus.INTERNAL_SERVER_ERROR (500).
     * @see HttpStatus
     */
    public HttpStatus getHttpStatus() {
        return httpStatus;
    }

    /**
     * Constructs a new runtime exception with the specified HttpStatus code and detail message.
     * The cause is not initialized, and may subsequently be initialized by a call to {@link #initCause}.
     * @param httpStatus the http status.  The detail message is saved for later retrieval by the {@link
     *                   #getHttpStatus()} method.
     * @param message    the detail message. The detail message is saved for later retrieval by the {@link
     *                   #getMessage()} method.
     * @see HttpStatus
     */
    public ResourceException(HttpStatus httpStatus, String message) {
        super(message);
        this.httpStatus = httpStatus;
    }
}

然后我创建一个控制器建议类

package com.javaninja.cam.spring;


import com.javaninja.cam.exception.ResourceException;

import org.springframework.http.ResponseEntity;
import org.springframework.web.bind.annotation.ExceptionHandler;


/**
 * Exception handler advice class for all SpringMVC controllers.
 * @author norrisshelton
 * @see org.springframework.web.bind.annotation.ControllerAdvice
 */
@org.springframework.web.bind.annotation.ControllerAdvice
public class ControllerAdvice {

    /**
     * Handles ResourceExceptions for the SpringMVC controllers.
     * @param e SpringMVC controller exception.
     * @return http response entity
     * @see ExceptionHandler
     */
    @ExceptionHandler(ResourceException.class)
    public ResponseEntity handleException(ResourceException e) {
        return ResponseEntity.status(e.getHttpStatus()).body(e.getMessage());
    }
}

使用它

throw new ResourceException(HttpStatus.BAD_REQUEST, "My message");

http://javaninja.net/2016/06/throwing-exceptions-messages-spring-mvc-controller/

于 2016-11-10T00:38:06.343 回答
11

我在我的 Spring Boot 应用程序中使用它

@RequestMapping(value = "/matches/{matchId}", produces = "application/json")
@ResponseBody
public ResponseEntity<?> match(@PathVariable String matchId, @RequestBody String body,
            HttpServletRequest request, HttpServletResponse response) {

    Product p;
    try {
      p = service.getProduct(request.getProductId());
    } catch(Exception ex) {
       return new ResponseEntity<String>(HttpStatus.BAD_REQUEST);
    }

    return new ResponseEntity(p, HttpStatus.OK);
}
于 2015-10-13T12:45:48.480 回答
3

使用 Spring Boot,我不完全确定为什么这是必要的(/error即使@ResponseBody在 an 上定义了我也得到了回退@ExceptionHandler),但以下内容本身不起作用:

@ResponseBody
@ResponseStatus(HttpStatus.BAD_REQUEST)
@ExceptionHandler(IllegalArgumentException.class)
public ErrorMessage handleIllegalArguments(HttpServletRequest httpServletRequest, IllegalArgumentException e) {
    log.error("Illegal arguments received.", e);
    ErrorMessage errorMessage = new ErrorMessage();
    errorMessage.code = 400;
    errorMessage.message = e.getMessage();
    return errorMessage;
}

它仍然抛出异常,显然是因为没有可生产的媒体类型被定义为请求属性:

// AbstractMessageConverterMethodProcessor
@SuppressWarnings("unchecked")
protected <T> void writeWithMessageConverters(T value, MethodParameter returnType,
        ServletServerHttpRequest inputMessage, ServletServerHttpResponse outputMessage)
        throws IOException, HttpMediaTypeNotAcceptableException, HttpMessageNotWritableException {

    Class<?> valueType = getReturnValueType(value, returnType);
    Type declaredType = getGenericType(returnType);
    HttpServletRequest request = inputMessage.getServletRequest();
    List<MediaType> requestedMediaTypes = getAcceptableMediaTypes(request);
    List<MediaType> producibleMediaTypes = getProducibleMediaTypes(request, valueType, declaredType);
if (value != null && producibleMediaTypes.isEmpty()) {
        throw new IllegalArgumentException("No converter found for return value of type: " + valueType);   // <-- throws
    }

// ....

@SuppressWarnings("unchecked")
protected List<MediaType> getProducibleMediaTypes(HttpServletRequest request, Class<?> valueClass, Type declaredType) {
    Set<MediaType> mediaTypes = (Set<MediaType>) request.getAttribute(HandlerMapping.PRODUCIBLE_MEDIA_TYPES_ATTRIBUTE);
    if (!CollectionUtils.isEmpty(mediaTypes)) {
        return new ArrayList<MediaType>(mediaTypes);

所以我添加了它们。

@ResponseBody
@ResponseStatus(HttpStatus.BAD_REQUEST)
@ExceptionHandler(IllegalArgumentException.class)
public ErrorMessage handleIllegalArguments(HttpServletRequest httpServletRequest, IllegalArgumentException e) {
    Set<MediaType> mediaTypes = new HashSet<>();
    mediaTypes.add(MediaType.APPLICATION_JSON_UTF8);
    httpServletRequest.setAttribute(HandlerMapping.PRODUCIBLE_MEDIA_TYPES_ATTRIBUTE, mediaTypes);
    log.error("Illegal arguments received.", e);
    ErrorMessage errorMessage = new ErrorMessage();
    errorMessage.code = 400;
    errorMessage.message = e.getMessage();
    return errorMessage;
}

这让我通过了“支持的兼容媒体类型”,但它仍然无法正常工作,因为我ErrorMessage的错误:

public class ErrorMessage {
    int code;

    String message;
}

JacksonMapper 没有将其作为“可转换”处理,所以我不得不添加 getter/setter,并且我还添加了@JsonProperty注解

public class ErrorMessage {
    @JsonProperty("code")
    private int code;

    @JsonProperty("message")
    private String message;

    public int getCode() {
        return code;
    }

    public void setCode(int code) {
        this.code = code;
    }

    public String getMessage() {
        return message;
    }

    public void setMessage(String message) {
        this.message = message;
    }
}

然后我按预期收到了我的消息

{"code":400,"message":"An \"url\" parameter must be defined."}
于 2016-12-01T15:41:05.693 回答
2

另一种方法是使用@ExceptionHandlerwith@ControllerAdvice将所有处理程序集中在同一个类中,如果不是,则必须将处理程序方法放在要管理异常的每个控制器中。

您的处理程序类:

@ControllerAdvice
public class MyExceptionHandler extends ResponseEntityExceptionHandler {

  @ExceptionHandler(MyBadRequestException.class)
  public ResponseEntity<MyError> handleException(MyBadRequestException e) {
    return ResponseEntity
        .badRequest()
        .body(new MyError(HttpStatus.BAD_REQUEST, e.getDescription()));
  }
}

您的自定义异常:

public class MyBadRequestException extends RuntimeException {

  private String description;

  public MyBadRequestException(String description) {
    this.description = description;
  }

  public String getDescription() {
    return this.description;
  }
}

现在您可以从任何控制器抛出异常,并且可以在您的建议类中定义其他处理程序。

于 2020-01-16T15:21:27.350 回答
0

您也可以throw new HttpMessageNotReadableException("error description")从 Spring 的默认错误处理中受益。

但是,就像那些默认错误一样,不会设置响应正文。

我发现这些在拒绝可能仅是手工制作的请求时很有用,这可能表明存在恶意意图,因为它们掩盖了基于更深层次的自定义验证及其标准而拒绝请求的事实。

Hth, dtk

于 2019-07-09T21:57:24.127 回答
0

在控制器中处理异常而无需显式返回的最简单和最干净的方法ResponseEntity就是添加@ExceptionHandler方法。

使用 Spring Boot 2.0.3.RELEASE 的示例片段:

// Prefer static import of HttpStatus constants as it's cleaner IMHO

// Handle with no content returned
@ExceptionHandler(IllegalArgumentException.class)
@ResponseStatus(BAD_REQUEST)
void onIllegalArgumentException() {}

// Return 404 when JdbcTemplate does not return a single row
@ExceptionHandler(IncorrectResultSizeDataAccessException.class)
@ResponseStatus(NOT_FOUND)
void onIncorrectResultSizeDataAccessException() {}

// Catch all handler with the exception as content
@ExceptionHandler(Exception.class)
@ResponseStatus(I_AM_A_TEAPOT)
@ResponseBody Exception onException(Exception e) {
  return e;
}

作为旁白:

  • 如果在所有上下文/用法matchService.getMatchJson(matchId) == null中都无效,那么我的建议是getMatchJson抛出异常,例如IllegalArgumentException而不是返回null并让它冒泡到控制器的@ExceptionHandler.
  • 如果null用于测试其他条件,那么我将有一个特定的方法,例如matchService.hasMatchJson(matchId)。一般来说,我null会尽可能避免以避免意外NullPointerException
于 2021-12-09T00:08:57.920 回答
-1

我认为这个线程实际上有最简单、最干净的解决方案,它不会牺牲 Spring 提供的 JSON 军事工具:

https://stackoverflow.com/a/16986372/1278921

于 2014-05-27T04:40:07.280 回答
-1

带有状态码的自定义响应

像那样:

class Response<T>(
        val timestamp: String = DateTimeFormatter
                .ofPattern("yyyy-MM-dd HH:mm:ss.SSSSSS")
                .withZone(ZoneOffset.UTC)
                .format(Instant.now()),
        val code: Int = ResultCode.SUCCESS.code,
        val message: String? = ResultCode.SUCCESS.message,
        val status: HttpStatus = HttpStatus.OK,
        val error: String? = "",
        val token: String? = null,
        val data: T? = null
) : : ResponseEntity<Response.CustomResponseBody>(status) {

data class CustomResponseBody(
        val timestamp: String = DateTimeFormatter
                .ofPattern("yyyy-MM-dd HH:mm:ss.SSSSSS")
                .withZone(ZoneOffset.UTC)
                .format(Instant.now()),
        val code: Int = ResultCode.SUCCESS.code,
        val message: String? = ResultCode.SUCCESS.message,
        val error: String? = "",
        val token: String? = null,
        val data: Any? = null
)

override fun getBody(): CustomResponseBody? = CustomResponseBody(timestamp, code, message, error, token, data)
于 2021-06-28T08:25:34.420 回答