java - 返回REST API的漂亮错误JSON

Question

基于Java的Rest Web服务中发生错误时
我收到这样的异常发送给客户端

 type Exception report

message Invalid Token

description The server encountered an internal error that prevented it from fulfilling this request.

exception

org.springframework.security.authentication.AuthenticationServiceException: Invalid Token
    com.resource.security.TokenAuthenticationFilter.doFilter(TokenAuthenticationFilter.java:220)
    org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    org.springframework.security.web.authentication.logout.LogoutFilter.doFilter(LogoutFilter.java:110)
    org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    org.springframework.security.web.context.request.async.WebAsyncManagerIntegrationFilter.doFilterInternal(WebAsyncManagerIntegrationFilter.java:50)
    org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:108)
    org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter(SecurityContextPersistenceFilter.java:87)
    org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    org.springframework.security.web.FilterChainProxy.doFilterInternal(FilterChainProxy.java:192)
    org.springframework.security.web.FilterChainProxy.doFilter(FilterChainProxy.java:160)
    org.springframework.web.filter.DelegatingFilterProxy.invokeDelegate(DelegatingFilterProxy.java:344)
    org.springframework.web.filter.DelegatingFilterProxy.doFilter(DelegatingFilterProxy.java:261)
    com.thetransactioncompany.cors.CORSFilter.doFilter(CORSFilter.java:208)
    com.thetransactioncompany.cors.CORSFilter.doFilter(CORSFilter.java:271)
note The full stack trace of the root cause is available in the Apache Tomcat/8.0.23 logs.

但我想返回这样的回复

{
"code" : 500,
"message" : "invalid token"
}

如何才能做到这一点 ？

更新

@Provider
   public class MyApplicationExceptionHandler implements
    ExceptionMapper<WebApplicationException> {

@Override
public Response toResponse(WebApplicationException weException) {

    // get initial response
    Response response = weException.getResponse();
    // create custom error
    ErrorDTO error = new ErrorDTO();
    error.setCode(response.getStatus());
    error.setMessage(weException.getMessage());
    // return the custom error
    return Response.status(response.getStatus()).entity(error).build();
}

}

Web.xml

  <context-param>
       <param-name>resteasy.providers</param-name>
       <param-value>com.madzz.common.exception.MadzzApplicationExceptionHandler</param-value>
 </context-param>

应用程序代码:

public String getTrackingDetailById(long orderItemId) throws Exception {
 throw new NotFoundException("not found"); }

我正在使用java.ws.rs.NotFoundException。但这似乎不起作用。
任何指针为什么？

Answer 1

您正在寻找的是@ControllerAdvice。逻辑如下:

您可以创建一个带有注释的类，其中该类中的每个方法都会响应一个或多个异常。这里的例子:

        @ControllerAdvice
    public class MyExceptionHandler {

        private static final Logger logger = LoggerFactory.getLogger(MyExceptionHandler.class);
        @ExceptionHandler(MyCustomException.class)
@ResponseBody
        public ExcObject handleSQLException(HttpServletRequest request, Exception ex){
            logger.info("SQLException Occured:: URL="+request.getRequestURL());
            return "database_error";
        }

        @ResponseStatus(value=HttpStatus.NOT_FOUND, reason="IOException occured")
        @ExceptionHandler(IOException.class)
        public void handleIOException(){
            logger.error("IOException handler executed");
            //returning 404 error code
        }
    }

在handleSQLException内部构造新创建的类ExcObject的新创建的对象，并返回该对象。

在您的 Controller 中，您需要引发特定的异常。

还请注意，您需要创建MyCustomException，它将扩展异常。