我使用Spring MVC为一个简单的JSON API,使用@ResponseBody的方法,如下所示。(我已经有了一个直接生成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;
}

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?

我确实遇到过这样的方法:

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。


当前回答

使用带有状态代码的自定义响应。

是这样的:

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)

其他回答

正如在一些回答中提到的,可以为您想要返回的每个HTTP状态创建一个异常类。我不喜欢为每个项目的每个状态创建一个类的想法。这是我想出的替代方案。

创建接受HTTP状态的通用异常 创建一个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通知类

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/

这不是最紧凑的方式,但在我看来很干净:

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

@ExceptionHandler(BadThingException.class)
@ResponseStatus(value = HttpStatus.BAD_REQUEST)
public @ResponseBody MyError handleException(BadThingException e) {
    return new MyError("That doesn’t work");
}

如果使用Spring 3.1+,您可以在异常处理方法中使用@ResponseBody,否则使用ModelAndView或其他东西。

@ResponseBody不能与@ExceptionHandler [sp -6902] #11567一起工作

您还可以抛出新的HttpMessageNotReadableException(“错误描述”)来受益于Spring的默认错误处理。

但是,与那些默认错误的情况一样,不会设置响应体。

我发现这些在拒绝合理地只能手工制作的请求(可能表明恶意意图)时很有用,因为它们掩盖了基于更深层次的自定义验证及其标准拒绝请求的事实。

我在Spring Boot应用程序中使用这个:

@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);
}

最简单的方法是抛出一个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;
}