URL to invoke Spring webservice

[Kunal Sathe]

Hello Folks,

I am relatively new to Spring WebServices and would like to know what url to use to invoke my webservice. The wsdl is auto generated and I have attached my spring-ws-servlet.xml file below. Any help will be highly appreciated.

Thanks,
Kunal

   <?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans"
	   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	   xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd">


    <bean id="rpmEndpoint" class="com.mccamish.event.service.RPMEndpoint">
        <constructor-arg ref="RPMService"/>
    </bean>

    <bean id="RPMService" class="com.mccamish.event.service.RPMService"/>
    
    <bean class="org.springframework.ws.server.endpoint.mapping.PayloadRootQNameEndpointMapping">
	    <property name="mappings">
	        <props>
	            <prop key="{http://com.mccamish.services}ServiceCallRequest">rpmEndpoint</prop>
	        </props>
	    </property>
	    <property name="interceptors">
	        <bean class="org.springframework.ws.server.endpoint.interceptor.PayloadLoggingInterceptor"/>
	    </property>
	</bean>
	
	<bean id="rpm" class="org.springframework.ws.wsdl.wsdl11.DefaultWsdl11Definition">
  		<property name="schema" ref="schema"/>
  		<property name="portTypeName" value="rpm"/>
	    <property name="locationUri" value="http://swdlvapp171:8280/RPMService/"/>
	</bean>

	<bean id="schema" class="org.springframework.xml.xsd.SimpleXsdSchema">
	    <property name="xsd" value="/WEB-INF/rpm.xsd"/>	    
	</bean>	
		
</beans> 

[Ivan Krizsan]

Hi!
Have you tried accessing the WSDL at http://swdlvapp171:8280/RPMService/rpm.wsdl
?
In the WSDL, you will hopefully find the address at which to invoke the service.
Best wishes!

[Kunal Sathe]

Thanks for the response Ivan.

I changed the locationuri to /RPMService/
Made it a relative path.

I am now able to access the wsdl and that does contain the location.

Thanks for your help again!

Kunal