Hi,
i am using WSSE with authentication and encryption.
Therefore I configured my clientApplication.xml
<bean id="clientFactory" class="org.apache.cxf.jaxws.JaxWsProxyFactoryBean">
<property name="serviceClass" value="demo.spring.ProductServiceImpl" />
<property name="outInterceptors">
<list>
<bean
class="org.apache.cxf.binding.soap.saaj.SAAJOutInterceptor" />
<ref bean="wss4jOutConfiguration" />
</list>
</property>
<property name="inInterceptors">
<list>
<bean
class="org.apache.cxf.binding.soap.saaj.SAAJInInterceptor" />
<ref bean="wss4jInConfiguration" />
</list>
</property>
</bean>
When I use wsdl2java I get a client.
@WebServiceClient(name = "ProductServicesBLA",
wsdlLocation =
"file:/Users/karlschmeing/Documents/workspace/cxf-webservices/src/b2b-bla-productservicesbla.wsdl",
targetNamespace =
"http://www.server.de/b2b/contract/bla/productservicesbla_v01_00")
public class ProductServicesBLA_Service extends Service {
public final static URL WSDL_LOCATION;
public final static QName SERVICE = new
QName("http://www.server.de/locationname", "ProductServicesBLA");
public final static QName ProductServicesBLAPort = new
QName("http://http://www.server.de/locationname", "ProductServicesBLA");
static {
URL url = null;
try {
url = new
URL("file:/Users/karlschmeing/Documents/workspace/cxf-webservices/src/b2b-bla-productservicestbla.wsdl");
} catch (MalformedURLException e) {
System.err.println("Can not initialize the default wsdl from
file:/Users/karlschmeing/Documents/workspace/cxf-webservices/src/b2b-bla-productservicesbla.wsdl");
// e.printStackTrace();
}
WSDL_LOCATION = url;
}
public ProductServicesBLA_Service(URL wsdlLocation) {
super(wsdlLocation, SERVICE);
}
How can i integrate the WebServiceClient into the Spring configuration? Is it
possible to annotate the in and out interceptor?
Regards,
Ralf Josephy