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1. Re: How to use a HTTP(S) proxy without system properties?
ron_sigal Aug 20, 2011 11:13 PM (in response to rspielmann)1 of 1 people found this helpfulHi Robert,
re: "but how would I be able to hook into that invocation?"
What is the nature of your client? I don't know much about web services, so I don't know what you mean by "webservice client generated by CXF". But I can say that if you don't have direct access to the actual Remoting call Client.invoke(), I don't see how will be able to set the value of the metadata map.
-Ron
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2. Re: How to use a HTTP(S) proxy without system properties?
rspielmann Aug 21, 2011 9:17 AM (in response to ron_sigal)Hi Ron,
"webservice client generated by CXF" means that I used a tool from the Apache CXF framework in order to generate required Java classes, including some client code, from the given WSDL. I then rely on the JBoss webservice stack to send the request. You are absolutely right: I was able to step through the entire stack in the debugger and found out that no matter which properties I set in the metadata map while I still have access to the request being prepared (there are some properties for proxy options), they are discarded shortly before the request is sent out to the target endpoint. Thus, I could not programmatically influence the proxy to be used. JBoss even ignored system properties, so I was pretty lost. In the end, the case was solved because I could use a transparent proxy by changing the host name part of the endpoint, so I didn't have to fiddle around with proxy settings any more. Thanks a lot for your reply though, it does point in the right direction
Regards,
Robert
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3. Re: How to use a HTTP(S) proxy without system properties?
ron_sigal Aug 22, 2011 3:05 PM (in response to rspielmann)Hey Robert,
"webservice client generated by CXF" means that I used a tool from the Apache CXF framework in order to generate required Java classes, including some client code, from the given WSDL.
Ah, yeah, should have guessed that. Thanks.
Anyway, I'm glad you got things working.
-Ron