如何在apache的spring流中获取HttpServletRequest

Question

我有一个简单的Apache Camel EIP流，其中入口点是spring服务。如何访问服务请求的HttpServletRequest对象？

from("spring-ws:uri:http://localhost:8080/test-ws-1.0/ws/TestService?endpointMapping=#endpointMapping")
.log("body is:\n${body}")
.process(new HttpRequestParserProcessor())

我希望能够从HttpRequestParserProcessor中获取请求的用户主体名称和请求的远程IP地址。调试处理器时，Exchange的in对象显示为SpringWebserviceMessage对象。

提前谢谢你，下午

Answer 1

来自http://docs.spring.io/spring-ws/sites/1.5/reference/html/common.html

TransportContext context = TransportContextHolder.getTransportContext();
HttpServletConnection connection = (HttpServletConnection )context.getConnection();
HttpServletRequest request = connection.getHttpServletRequest();
String ipAddress = request.getRemoteAddr();

或者使用CXF代替Spring。CXF来自与骆驼相同的厨房，因此可能会更好地整合。使用CXF，您可以访问ServletRequest，如下所示(参见这里)：

org.apache.cxf.message.Message cxfMessage = in.getHeader(CxfConstants.CAMEL_CXF_MESSAGE, org.apache.cxf.message.Message.class);
ServletRequest request = (ServletRequest)cxfMessage.get("HTTP.REQUEST");