Hi,
Ensure the
bus.setExtension(myHttpConduitConfig, MyHTTPConduitConfigurer.class);
before
Client client = dcf.createClient("http://example.com/EchoService?wsdl";);
－－－－－－－－－－－－－
Freeman(Yue) Fang
Red Hat, Inc.
FuseSource is now part of Red Hat
Web: http://fusesource.com | http://www.redhat.com/
Twitter: freemanfang
Blog: http://freemanfang.blogspot.com
http://blog.sina.com.cn/u/1473905042
weibo: @Freeman小屋
On 2013-3-14, at 下午11:08, Alos wrote:
> Thanks for all the help!
>
> I was unable to register the MyHTTPConduitConfigurer as there is no way to
> get a reference to the JaxWsDynamicClientFactory's bus. What I ended up doing
> was subclassing the JaxWsDynamicClientFactory and overwriting the method
> where the bus is created!
>
> However when I use my new subclass I never see my System.outs that I added to
> my MyHTTPConduitConfigurer.
>
> Any ideas on what might be happening?
>
> Sent from my iPhone
>
> On Mar 12, 2013, at 11:57 PM, Freeman Fang <[email protected]> wrote:
>
>> Hi,
>>
>> Yes, you can write your own HTTPConduitConfigurer and register to the bus
>> which you get fromJaxWsDynamicClientFactory name, String address,
>> HTTPConduit c) {
>> AuthorizationPolicy authorizationPolicy = new AuthorizationPolicy();
>> authorizationPolicy.setUserName("...");
>> authorizationPolicy.setPassword("...");
>> c.setAuthorization(authorizationPolicy);
>> }
>> }
>>
>> then register MyHTTPConduitConfigurer to the bus which you get from
>> JaxWsDynamicClientFactory, something like
>> MyHTTPConduitConfigurer myHttpConduitConfig = new MyHTTPConduitConfigurer();
>> bus.setExtension(myHttpConduitConfig, MyHTTPConduitConfigurer.class);
>>
>> Hope this helps
>> －－－－－－－－－－－－－
>> Freeman(Yue) Fang
>>
>> Red Hat, Inc.
>> FuseSource is now part of Red Hat
>> Web: http://fusesource.com | http://www.redhat.com/
>> Twitter: freemanfang
>> Blog: http://freemanfang.blogspot.com
>> http://blog.sina.com.cn/u/1473905042
>> weibo: @Freeman小屋
>>
>> On 2013-3-13, at 上午2:04, Alos wrote:
>>
>>> This would force me to have a "hard coded" XML with http:conduits for each
>>> of the services using this authentication and thus defeating the purpose of
>>> my DynamicClient. Is there a way of creating these http:conduits in code?
>>>
>>> BTW thanks so much for responding so quickly :-)
>>>
>>> Sent from my iPhone
>>>
>>> On Mar 11, 2013, at 8:40 PM, Freeman Fang <[email protected]> wrote:
>>>
>>>> Hi,
>>>>
>>>> You need add a spring configuration, add something like
>>>>
>>>> <http:conduit name="http://example.com/.*";>
>>>> <authorization>
>>>> <sec:UserName>myuser</sec:UserName>
>>>> <sec:Password>mypasswd</sec:Password>
>>>> <sec:AuthorizationType>Basic</sec:AuthorizationType>
>>>> </authorization>
>>>> </http:conduit>
>>>>
>>>> take a look at [1] to get more details
>>>>
>>>> [1]http://cxf.apache.org/docs/client-http-transport-including-ssl-support.html
>>>>
>>>> －－－－－－－－－－－－－
>>>> Freeman(Yue) Fang
>>>>
>>>> Red Hat, Inc.
>>>> FuseSource is now part of Red Hat
>>>> Web: http://fusesource.com | http://www.redhat.com/
>>>> Twitter: freemanfang
>>>> Blog: http://freemanfang.blogspot.com
>>>> http://blog.sina.com.cn/u/1473905042
>>>> weibo: @Freeman小屋
>>>>
>>>> On 2013-3-12, at 上午8:09, alos wrote:
>>>>
>>>>> Hey guys!
>>>>>
>>>>> So I've been working with the JaxWsDynamicClientFactory. I´m always
>>>>> able to get the WSDL like this:
>>>>>
>>>>> JaxWsDynamicClientFactory dcf = JaxWsDynamicClientFactory.newInstance();
>>>>> Client client = dcf.createClient("http://example.com/EchoService?wsdl";);
>>>>>
>>>>> However a client's new WSDL requires me to pass along an
>>>>> authentication header in the HTTP message.
>>>>>
>>>>> Is there a way to do this using the JaxWsDynamicClientFactory??
>>>>>
>>>>> So far we´ve tried to get a hold of an Interceptor (so I can get the
>>>>> HTTPRequest object and do my thing) but it looks like these are
>>>>> created AFTER the client is instanced.
>>>>>
>>>>> Any help will be appreciated!
>>
