如何从 Spring 中的 WebRequest 获取请求的 URI？

Question

我正在 spring Rest 网络服务中使用 @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 class:

public class ApiError {

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

但是WebRequest没有给出任何api来获取请求失败的路径。我试过： request.toString() returns -> ServletWebRequest: uri=/signup;client=0:0:0:0:0:0:0:1
request.getDescription(false) returns -> 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

找到解决方案。将 WebRequest 转换为 ServletWebRequest 解决了这个问题。

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

returns 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 获取请求 URI 和客户端信息 webRequest.getDescription(真).

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 而不是 WebRequest as

@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 @ControllerAdvice class 的方便基础 class 希望在所有 @RequestMapping 中提供集中式异常处理通过@ExceptionHandler 方法的方法。 here

在SpringBoot 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

您可以使用 request.getDescription(false)。

Answer 6

我正在使用 SpringBoot 2.5.3 和 globalExceptionHandler。简短的片段。使用“TheCoder”答案并从那里开始。如果不需要，则不必使用 header、status、... WebRequest 作为输入参数。这只给出了 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();
}

如何从 Spring 中的 WebRequest 获取请求的 URI？

How to get request's URI from WebRequest in Spring?

spring

spring-mvc

spring-rest