[英]MessageContext in SOAPHandler JAX-WS WebService
I have a JAX-WS 2.2 WebService and I must take the IP Address of each client that communicate with it. 我有一个JAX-WS 2.2 WebService,我必须获取与之通信的每个客户端的IP地址。 I write a SOAP protocol handler but I can't see the addresses because the handlers doesn't contain this information and using the mimeheaders I can't also see this information. 我写了一个SOAP协议处理程序,但我看不到地址,因为处理程序不包含这些信息,并且使用mimeheaders我也看不到这些信息。 The code of my handler is the follow: 我的处理程序的代码如下:
public class AddressHandler implements SOAPHandler<SOAPMessageContext> {
private void takeIPAddress(SOAPMessageContext context) {
try {
SOAPMessage original = context.getMessage();
MimeHeaders mimeheaders = original.getMimeHeaders();
MimeHeader mimeheader = null;
Iterator<?> iter = mimeheaders.getAllHeaders();
for (; iter.hasNext();) {
mimeheader = (MimeHeader) iter.next();
System.out.println("name=" + mimeheader.getName() + ", value="
+ mimeheader.getValue());
}
} catch (Exception e) {
e.printStackTrace();
}
}
@Override
public void close(MessageContext arg0) {
// TODO Auto-generated method stub
}
@Override
public boolean handleFault(SOAPMessageContext arg0) {
// TODO Auto-generated method stub
return false;
}
@Override
public boolean handleMessage(SOAPMessageContext context) {
takeIPAddress(context);
return true;
}
@Override
public Set<QName> getHeaders() {
// TODO Auto-generated method stub
return null;
}
}
Now I'm seeing that would be possible to see the addresses using the following code: 现在我看到可以使用以下代码查看地址:
SOAPMessageContext jaxwsContext = (SOAPMessageContext)wsContext.getMessageContext();
HttpServletRequest request = HttpServletRequest)jaxwsContext.get(SOAPMessageContext.SERVLET_REQUEST);
String ipAddress = request.getRemoteAddr();
But I can't import correctly the HttpServletRequest class. 但我无法正确导入HttpServletRequest类。 Do you have any ideas? 你有什么想法?
UPDATE UPDATE
Thanks to A Nyar Thar, I've seen that exists another method to take address and I've implemented this in my code, that now is: 感谢A Nyar Thar,我已经看到存在另一种获取地址的方法,我已经在我的代码中实现了这个,现在是:
private void takeIPAddress(SOAPMessageContext context) {
HttpExchange exchange = (HttpExchange)context.get("com.sun.xml.ws.http.exchange");
InetSocketAddress remoteAddress = exchange.getRemoteAddress();
String remoteHost = remoteAddress.getHostName();
System.out.println(remoteHost);
}
But the code execution create this error (where row 39 is where I do exchange.getRemoteAddress()): 但代码执行会产生此错误(其中第39行是我执行exchange.getRemoteAddress()的地方):
java.lang.NullPointerException
at server.AddressHandler.takeIPAddress(AddressHandler.java:39)
at server.AddressHandler.handleMessage(AddressHandler.java:80)
at server.AddressHandler.handleMessage(AddressHandler.java:1)
at com.sun.xml.internal.ws.handler.HandlerProcessor.callHandleMessage(HandlerProcessor.java:282)
at com.sun.xml.internal.ws.handler.HandlerProcessor.callHandlersRequest(HandlerProcessor.java:125)
at com.sun.xml.internal.ws.handler.ServerSOAPHandlerTube.callHandlersOnRequest(ServerSOAPHandlerTube.java:123)
at com.sun.xml.internal.ws.handler.HandlerTube.processRequest(HandlerTube.java:105)
at com.sun.xml.internal.ws.api.pipe.Fiber.__doRun(Fiber.java:626)
at com.sun.xml.internal.ws.api.pipe.Fiber._doRun(Fiber.java:585)
at com.sun.xml.internal.ws.api.pipe.Fiber.doRun(Fiber.java:570)
at com.sun.xml.internal.ws.api.pipe.Fiber.runSync(Fiber.java:467)
at com.sun.xml.internal.ws.server.WSEndpointImpl$2.process(WSEndpointImpl.java:299)
at com.sun.xml.internal.ws.transport.http.HttpAdapter$HttpToolkit.handle(HttpAdapter.java:593)
at com.sun.xml.internal.ws.transport.http.HttpAdapter.handle(HttpAdapter.java:244)
at com.sun.xml.internal.ws.transport.http.server.WSHttpHandler.handleExchange(WSHttpHandler.java:95)
at com.sun.xml.internal.ws.transport.http.server.WSHttpHandler.handle(WSHttpHandler.java:80)
at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:77)
at sun.net.httpserver.AuthFilter.doFilter(AuthFilter.java:83)
at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:80)
at sun.net.httpserver.ServerImpl$Exchange$LinkHandler.handle(ServerImpl.java:677)
at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:77)
at sun.net.httpserver.ServerImpl$Exchange.run(ServerImpl.java:649)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at java.lang.Thread.run(Thread.java:744)
I think that the real problem is that I don't know how take WebServiceContext from my class AddressHandler. 我认为真正的问题是我不知道如何从我的类AddressHandler中获取WebServiceContext。 Do you have ideas? 你有想法吗?
For JAX-WS based webservice, you can access remote host info from javax.xml.ws.spi.http.HttpExchange
, that can be accessed based on JAX-WS
version, 对于基于JAX-WS的Web服务,您可以从javax.xml.ws.spi.http.HttpExchange
访问远程主机信息,可以基于JAX-WS
版本访问,
SOAPMessageContext soapContext = (SOAPMessageContext)wsContext.getMessageContext(); HttpExchange exchange = (HttpExchange)soapContext.get(JAXWSProperties.HTTP_EXCHANGE);
SOAPMessageContext soapContext = (SOAPMessageContext)wsContext.getMessageContext(); HttpExchange exchange = (HttpExchange)soapContext.get("com.sun.xml.ws.http.exchange");
Note that wsContext.getMessageContext()
will return MessageContext
. 请注意, wsContext.getMessageContext()
将返回MessageContext
。 If you want it, don't cast to SOAPMessageContext
, like that, 如果你想要它,不要像那样SOAPMessageContext
为SOAPMessageContext
,
MessageContext msgContext = wsContext.getMessageContext();
Finally you can access remote address info, 最后,您可以访问远程地址信息,
InetSocketAddress remoteAddress = exchange.getRemoteAddress();
String remoteHost = remoteAddress.getHostName();
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