这样的方法应该是可行的，但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;
}

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/

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)

将您的返回类型更改为ResponseEntity<>，然后您可以使用下面的400:

return new ResponseEntity<>(HttpStatus.BAD_REQUEST);

对于正确的请求:

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

Spring 4.1之后，ResponseEntity中有了帮助方法，可以用作:

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

and

return ResponseEntity.ok(json);

这样的方法应该是可行的，但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;
}

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;
}