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  • XSD import in WSDL using relative path

    I am using Spring 3.0.5.
    I have an existing wsdl file let's name it a.wsdl
    I have imported in a.wsdl the a.xsd xml schema assuming relative path.
    a.wsdl and a.xsd is under <WebRoot>/WEB-INF/wsdl directory.

    How do I expose the a.xsd so it will be recognize and not having a HTTP response code of 405 "The specified HTTP method is not allowed for the requested resource"

    I am trying to use soapUI and has this exception whenever I am using the url of the WSDL. for example http://localhost:8080/someWebContext/a.wsdl

    I am using Apache Tomcat 6.0.20. Is this just a problem in Web server?

    My web.xml has the following
    Code:
        
    <servlet>
      <servlet-name>spring-ws</servlet-name>
      <servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
      <init-param>
        <!-- Transform the location attributes in WSDLs -->
        <param-name>transformWsdlLocations</param-name>
        <param-value>true</param-value>
      </init-param>
    </servlet>
    
    <servlet-mapping>
      <servlet-name>spring-ws</servlet-name>
      <url-pattern>/*</url-pattern>
    </servlet-mapping>
    I have the following for spring-ws-servlet.xml
    Code:
    <bean id="aXmlSchema" class="org.springframework.xml.xsd.SimpleXsdSchema">
      <property name="xsd" value="/WEB-INF/wsdl/a.xsd"/>
    </bean>
    <bean id="aWsdl" class="org.springframework.ws.wsdl.wsdl11.SimpleWsdl11Definition">
      <property name="wsdl" value="/WEB-INF/wsdl/a.wsdl"/>
    </bean>
    Putting the wsdl and xsd file out of the WEB-INF directory is out of the question. It is a must that the wsdl and xsd file resides under WEB-INF/wsdl directory.

    I have added SimpleXsdSchema bean entries for a.xsd but it didn't work. I am thinking the problem probably is inside the interpretation of wsdl file using SimpleWsdl11Definition?

    Does anyone have found a way to get around with my above problem?
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