Spring MVC REST通过返回JSON处理错误的网址（404）

  • Z时代
  • 2024-01-10
  • 分类：问答

我正在使用SpringMVC开发REST服务，其中在类和方法级别使用@RequestMapping。

当前已将该应用程序配置为返回在web.xml中配置的错误页面jsp。

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

但是，我想返回自定义JSON而不是此错误页面。

通过在控制器中编写此代码，我能够处理异常并为其他异常返回json，但不确定在URL根本不存在时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);
        }

@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);
    }
}

回答：

在SpringFramework中研究了DispatcherServlet和HttpServletBean.init（）之后，我发现在Spring

4中它是可能的。

/** 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);
    }
}

为false，我们应该在 *

<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，您可以使用此方法将其捕获在带有 注释的类中。

@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);
}

这使我可以为不存在映射的错误URL返回JSON响应，而不是重定向到JSP页面:)

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

以上是 Spring MVC REST通过返回JSON处理错误的网址（404） 的全部内容， 来源链接： utcz.com/qa/419018.html

回到顶部