Spring MVC REST通过返回JSON处理坏URL（404）

Question

我正在使用SpringMVC开发一个REST服务，在那里我在类和方法级别有@刚需映射。

此应用程序当前配置为返回在web中配置的错误页面jsp。xml。

<error-page>
    <error-code>404</error-code>
    <location>/resourceNotFound</location>
</error-page>

但是，我希望返回自定义JSON，而不是此错误页面。

通过在controller中编写此代码，我能够处理异常并为其他异常返回json，但不确定在url根本不存在的情况下如何以及在何处编写返回json的逻辑。

    @ExceptionHandler(TypeMismatchException.class)
        @ResponseStatus(value=HttpStatus.NOT_FOUND)
        @ResponseBody
        public ResponseEntity<String> handleTypeMismatchException(HttpServletRequest req, TypeMismatchException ex) {

            HttpHeaders headers = new HttpHeaders();
            headers.add("Content-Type", "application/json; charset=utf-8");
            Locale locale = LocaleContextHolder.getLocale();
            String errorMessage = messageSource.getMessage("error.patient.bad.request", null, locale);

            errorMessage += ex.getValue();
            String errorURL = req.getRequestURL().toString();

            ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
            return new ResponseEntity<String>(errorInfo.toJson(), headers, HttpStatus.BAD_REQUEST);

        }

我试过@ControllerAdvice，它适用于其他异常情况，但不适用于映射不可用的情况，

@ControllerAdvice
public class RestExceptionProcessor {

    @Autowired
    private MessageSource messageSource;

    @ExceptionHandler(HttpRequestMethodNotSupportedException.class)
    @ResponseStatus(value=HttpStatus.NOT_FOUND)
    @ResponseBody
    public ResponseEntity<String> requestMethodNotSupported(HttpServletRequest req, HttpRequestMethodNotSupportedException ex) {
        Locale locale = LocaleContextHolder.getLocale();
        String errorMessage = messageSource.getMessage("error.patient.bad.id", null, locale);

        String errorURL = req.getRequestURL().toString();

        ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
        return new ResponseEntity<String>(errorInfo.toJson(), HttpStatus.BAD_REQUEST);
    }

    @ExceptionHandler(NoSuchRequestHandlingMethodException.class)
    @ResponseStatus(value=HttpStatus.NOT_FOUND)
    @ResponseBody
    public ResponseEntity<String> requestHandlingMethodNotSupported(HttpServletRequest req, NoSuchRequestHandlingMethodException ex) {
        Locale locale = LocaleContextHolder.getLocale();
        String errorMessage = messageSource.getMessage("error.patient.bad.id", null, locale);

        String errorURL = req.getRequestURL().toString();

        ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
        return new ResponseEntity<String>(errorInfo.toJson(), HttpStatus.BAD_REQUEST);
    }


}

Answer 1

如果您使用的是Spring 3.2或更高版本，您可以使用控制器建议（@ControlllerAdtions）来处理映射错误（404）等问题。你可以在这里找到留档。看看第17.11节。例如，您可以使用它来提供更详细的日志记录，说明为什么您的请求绑定没有与特定的网址匹配，或者简单地返回比通用404更具体的响应。

Answer 2

如果您使用的是Spring Boot，请同时设置以下两个属性：

spring.resources.add-mappings=false
spring.mvc.throw-exception-if-no-handler-found=true

现在@ControllerAdvice注释类可以处理“NoHandlerFoundException”，如下所示。

@ControllerAdvice
@RequestMapping(produces = "application/json")
@ResponseBody
public class RestControllerAdvice {

    @ExceptionHandler(NoHandlerFoundException.class)
    public ResponseEntity<Map<String, Object>> unhandledPath(final NoHandlerFoundException e) {
        Map<String, Object> errorInfo = new LinkedHashMap<>();
        errorInfo.put("timestamp", new Date());
        errorInfo.put("httpCode", HttpStatus.NOT_FOUND.value());
        errorInfo.put("httpStatus", HttpStatus.NOT_FOUND.getReasonPhrase());
        errorInfo.put("errorMessage", e.getMessage());
        return new ResponseEntity<Map<String, Object>>(errorInfo, HttpStatus.NOT_FOUND);
    }

}

注意：仅指定此属性是不够的：

spring.mvc.throw-exception-if-no-handler-found=true

，因为默认情况下Spring将未知URL映射到/**，所以实际上从来没有“找不到处理程序”。

要禁用到/**的未知url映射，您需要

spring.resources.add-mappings=false ,

这就是为什么这两个属性一起产生所需的行为。

Answer 3

在挖掘DispatcherServlet和HttpServletBean之后。SpringFramework中的init（）我发现它在Spring4中是可能的。

组织。springframework。网状物servlet。调度员服务

/** Throw a NoHandlerFoundException if no Handler was found to process this request? **/
private boolean throwExceptionIfNoHandlerFound = false;

protected void noHandlerFound(HttpServletRequest request, HttpServletResponse response) throws Exception {
    if (pageNotFoundLogger.isWarnEnabled()) {
        String requestUri = urlPathHelper.getRequestUri(request);
        pageNotFoundLogger.warn("No mapping found for HTTP request with URI [" + requestUri +
                "] in DispatcherServlet with name '" + getServletName() + "'");
    }
    if(throwExceptionIfNoHandlerFound) {
        ServletServerHttpRequest req = new ServletServerHttpRequest(request);
        throw new NoHandlerFoundException(req.getMethod().name(),
                req.getServletRequest().getRequestURI(),req.getHeaders());
    } else {
        response.sendError(HttpServletResponse.SC_NOT_FOUND);
    }
}

默认情况下，ThrowExceptionIfNoHandlerFIND为false，我们应该在web.xml中启用它

<servlet>
    <servlet-name>appServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>throwExceptionIfNoHandlerFound</param-name>
            <param-value>true</param-value>
        </init-param>
    <load-on-startup>1</load-on-startup>
    <async-supported>true</async-supported>
</servlet>

然后，您可以使用此方法在一个带有@ControlllerAdtions注释的类中捕获它。

@ExceptionHandler(NoHandlerFoundException.class)
@ResponseStatus(value=HttpStatus.NOT_FOUND)
@ResponseBody
public ResponseEntity<String> requestHandlingNoHandlerFound(HttpServletRequest req, NoHandlerFoundException ex) {
    Locale locale = LocaleContextHolder.getLocale();
    String errorMessage = messageSource.getMessage("error.bad.url", null, locale);

    String errorURL = req.getRequestURL().toString();

    ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
    return new ResponseEntity<String>(errorInfo.toJson(), HttpStatus.BAD_REQUEST);
}

它允许我返回没有映射存在的坏URL的JSON响应，而不是重定向到JSP页面：）

{"message":"URL does not exist","url":"http://localhost:8080/service/patientssd"}