如何在spring mvc @Controller中返回错误信息

Question

我正在使用这样的方法

@RequestMapping(method = RequestMethod.GET)
public ResponseEntity<UserWithPhoto> getUser(@RequestHeader(value="Access-key") String accessKey,
                                     @RequestHeader(value="Secret-key") String secretKey){
    try{
        return new ResponseEntity<UserWithPhoto>((UserWithPhoto)this.userService.chkCredentials(accessKey, secretKey, timestamp),
                new HttpHeaders(),
                HttpStatus.CREATED);
    }
    catch(ChekingCredentialsFailedException e){
        e.printStackTrace();
        return new ResponseEntity<UserWithPhoto>(null,new HttpHeaders(),HttpStatus.FORBIDDEN);
    }
}

我想在发生异常时返回一些文本消息,但现在我只返回status和null对象.有可能吗？

Answer 1

正如Sotirios Delimanolis在评论中已经指出的那样,有两种选择:

返回ResponseEntity错误消息

像这样改变你的方法:

@RequestMapping(method = RequestMethod.GET)
public ResponseEntity getUser(@RequestHeader(value="Access-key") String accessKey,
                              @RequestHeader(value="Secret-key") String secretKey) {
    try {
        // see note 1
        return ResponseEntity
            .status(HttpStatus.CREATED)                 
            .body(this.userService.chkCredentials(accessKey, secretKey, timestamp));
    }
    catch(ChekingCredentialsFailedException e) {
        e.printStackTrace(); // see note 2
        return ResponseEntity
            .status(HttpStatus.FORBIDDEN)
            .body("Error Message");
    }
}

注意1:您不必使用ResponseEntity构建器,但我发现它有助于保持代码可读.它还有助于记住特定HTTP状态代码的响应应包含哪些数据.例如,状态代码201的响应应该包含指向Location标题中新创建的资源的链接(请参阅状态代码定义).这就是Spring提供方便的构建方法的原因ResponseEntity.created(URI).

注意2:不要使用printStackTrace(),请改用记录器.

提供一个 @ExceptionHandler

从方法中删除try-catch块并让它抛出异常.然后在类注释的类中创建另一个方法@ControllerAdvice:

@ControllerAdvice
public class ExceptionHandlerAdvice {

    @ExceptionHandler(ChekingCredentialsFailedException.class)
    public ResponseEntity handleException(ChekingCredentialsFailedException e) {
        // log exception
        return ResponseEntity
                .status(HttpStatus.FORBIDDEN)
                .body("Error Message");
    }        
}

请注意,@ExceptionHandler允许注释的方法具有非常灵活的签名.有关详细信息,请参阅Javadoc.

Answer 2

这是另一种选择.创建一个带有状态代码和消息的通用异常.然后创建一个异常处理程序 使用异常处理程序从异常中检索信息并返回到服务的调用者.

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

public class ResourceException extends RuntimeException {

    private HttpStatus httpStatus = HttpStatus.INTERNAL_SERVER_ERROR;

    public HttpStatus getHttpStatus() {
        return httpStatus;
    }

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

然后使用异常处理程序检索信息并将其返回给服务调用者.

@ControllerAdvice
public class ExceptionHandlerAdvice { 

    @ExceptionHandler(ResourceException.class)
    public ResponseEntity handleException(ResourceException e) {
        // log exception 
        return ResponseEntity.status(e.getHttpStatus()).body(e.getMessage());
    }         
} 

然后在需要时创建一个例外.

throw new ResourceException(HttpStatus.NOT_FOUND, "We were unable to find the specified resource.");