如何在 Spring 中从 WebRequest 获取请求的 URI？

Question

我在 Spring Rest web 服务中使用 @ControllerAdvice 和 ResponseEntityExceptionHandler 处理 REST 异常。到目前为止一切正常，直到我决定将 URI 路径（发生异常）添加到 BAD_REQUEST 响应中。

@ControllerAdvice
public class RestResponseEntityExceptionHandler extends ResponseEntityExceptionHandler {

@Override
protected ResponseEntity<Object> handleHttpMessageNotReadable(HttpMessageNotReadableException ex,
        HttpHeaders headers, HttpStatus status, WebRequest request) {
    logger.info(request.toString());
    return handleExceptionInternal(ex, errorMessage(HttpStatus.BAD_REQUEST, ex, request), headers, HttpStatus.BAD_REQUEST, request);
}

private ApiError errorMessage(HttpStatus httpStatus, Exception ex, WebRequest request) {
    final String message = ex.getMessage() == null ? ex.getClass().getName() : ex.getMessage();
    final String developerMessage = ex.getCause() == null ? ex.toString() : ex.getCause().getMessage();
    return new ApiError(httpStatus.value(), message, developerMessage, System.currentTimeMillis(), request.getDescription(false));
}

ApiError 只是一个 Pojo 类：

public class ApiError {

    private Long timeStamp;
    private int status;
    private String message;
    private String developerMessage;
    private String path;
}

但是 WebRequest 并没有给出任何 api 来获取请求失败的路径。我试过了： request.toString() 返回 -> ServletWebRequest: uri=/signup;client=0:0:0:0:0:0:0:1
request.getDescription(false) 返回 -> uri =/signup
getDescription 非常接近要求，但不符合要求。有没有办法只获取 uri 部分？

Answer 1

找到了解决办法。将WebRequest 转换为ServletWebRequest 解决了目的。

((ServletWebRequest)request).getRequest().getRequestURI().toString()

返回完整路径 - http://localhost:8080/signup

Answer 2

这个问题有多种解决方案。

1) 可以使用 WebRequest 从 WebRequest 获取请求 URI 和客户端信息 webRequest.getDescription(true)。

true 将显示用户的信息，例如客户端 ID，false 将仅打印 URI。

2) 在方法定义中直接使用HttpServletRequest代替WebRequest作为

@Override
protected ResponseEntity<Object> handleHttpMessageNotReadable(HttpMessageNotReadableException ex,
        HttpHeaders headers, HttpStatus status, WebRequest request, HttpServletRequest httpRequest) {
    logger.info(httpRequest.getRequestURI());
    return handleExceptionInternal(ex, errorMessage(HttpStatus.BAD_REQUEST, ex, request), headers, HttpStatus.BAD_REQUEST, request);
}

Answer 3

ResponseEntityExceptionHandler 解释了一个方便的@ControllerAdvice 类基类，这些类希望通过@ExceptionHandler 方法在所有@RequestMapping 方法中提供集中的异常处理。 here

在 Spring Boot 2.1.6 中，可以这样写：

RestExceptionHandler.java

@Order(Ordered.HIGHEST_PRECEDENCE)
@RestControllerAdvice
public class RestExceptionHandler extends ResponseEntityExceptionHandler {
 
private static final Logger logger = LoggerFactory.getLogger(RestExceptionHandler.class);

@ExceptionHandler(ResourceNotFoundException.class)
protected ResponseEntity<Object> handleEntityNotFound(ResourceNotFoundException ex, final HttpServletRequest httpServletRequest) {
    ApiError apiError = new ApiError(HttpStatus.NOT_FOUND);
    apiError.setMessage("Resource not found");
    apiError.setDebugMessage(ex.getMessage());
    apiError.setPath(httpServletRequest.getRequestURI());
    return buildResponseEntity(apiError);
}

private ResponseEntity<Object> buildResponseEntity(ApiError apiError) {
    return new ResponseEntity<>(apiError, apiError.getStatus());
}


 
 @Override
protected ResponseEntity<Object> handleHttpRequestMethodNotSupported(HttpRequestMethodNotSupportedException ex, HttpHeaders headers, HttpStatus status, WebRequest request) {
        ApiError apiError = new ApiError(HttpStatus.METHOD_NOT_ALLOWED);
        apiError.setMessage(ex.getMessage());
        apiError.setPath(((ServletWebRequest)request).getRequest().getRequestURI().toString());
        logger.warn(ex.getMessage());
        return buildResponseEntity(apiError);
}
}

让我们从实现一个发送错误的简单结构开始：

ApiError.java

public class ApiError {
// 4xx and 5xx
private HttpStatus status;

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss.SSS'Z'")
private LocalDateTime timestamp;

// holds a user-friendly message about the error.
private String message;

// holds a system message describing the error in more detail.
@JsonInclude(value = Include.NON_EMPTY)
private String debugMessage;

// returns the part of this request's URL
private String path;

@JsonInclude(value = Include.NON_EMPTY)
private List<String> details=new ArrayList<>();

// setters & getters
}

ResourceNotFoundException.java

public class ResourceNotFoundException extends RuntimeException {

private static final long serialVersionUID = 1L;

public ResourceNotFoundException() {
    super();
}

public ResourceNotFoundException(String msg) {
    super(msg);
}

Answer 4

访问WebRequest对象的属性：

Object obj = webRequest.getAttribute("org.springframework.web.util.UrlPathHelper.PATH", 0)
String uri = String.valueOf(obj);

webRequest.getAttribute(String attributeName, int scope);

// scope can be either:
//   0: request
//   1: session

// valid attribute names can be fetched with call:
String[] attributeNames = webRequest.getAttributeNames(0);   //scope is request

有效的属性名称是：

org.springframework.web.util.UrlPathHelper.PATH
org.springframework.web.context.request.async.WebAsyncManager.WEB_ASYNC_MANAGER
org.springframework.web.servlet.HandlerMapping.bestMatchingHandler
org.springframework.web.servlet.DispatcherServlet.CONTEXT
org.springframework.web.servlet.resource.ResourceUrlProvider
characterEncodingFilter.FILTERED
org.springframework.boot.web.servlet.error.DefaultErrorAttributes.ERROR
org.springframework.web.servlet.DispatcherServlet.THEME_SOURCE
org.springframework.web.servlet.DispatcherServlet.LOCALE_RESOLVER
formContentFilter.FILTERED
org.springframework.web.servlet.HandlerMapping.bestMatchingPattern
requestContextFilter.FILTERED
org.springframework.web.servlet.DispatcherServlet.OUTPUT_FLASH_MAP
org.springframework.web.servlet.HandlerMapping.pathWithinHandlerMapping
org.springframework.web.servlet.DispatcherServlet.FLASH_MAP_MANAGER
org.springframework.web.servlet.HandlerMapping.uriTemplateVariables
org.springframework.web.servlet.DispatcherServlet.THEME_RESOLVER
org.springframework.core.convert.ConversionService

Answer 5

你可以使用 request.getDescription(false)。

Answer 6

我正在使用 SpringBoot 2.5.3 和 globalExceptionHandler。 短片。使用“TheCoder”答案并从那里开始。 如果您不需要它们，则不必使用标头、状态、... WebRequest 作为输入参数。这仅给出了 url 的端点，而不是主机名。

@ExceptionHandler(value = NotFound.class)
ResponseEntity<...> httpNotFoundException(NotFound exc, HttpServletRequest req ) {
//use req.getRequestURI();
}

@ExceptionHandler(value = HttpClientErrorException.class)
ResponseEntity<...> httpClientException(HttpClientErrorException exc, HttpServletRequest req ) {
exc.getRawStatusCode()  //to get status code
//I am using this to check for 404 and handling here with other stuff instead of using NotFound.class above.
// Use req.getRequestURI();
}