Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
162 views
in Technique[技术] by (71.8m points)

java - accessing wsdl on Tomcat

I have a web service and I was deploying it on GlassFish. I accessed its wsdl through http://localhost:10697/APIService/APIServiceService?wsdl.

Now I ported the WAR file to a Tomcat 6.0.24 and it is deployed. However I am trying to access its wsdl using http://localhost:8080/APIService/APIServiceService?wsdl but I'm getting a 404 error. I tried various combinations but none seem to work.

How can I access the wsdl file plz?

Thanks and regards,

Update: Here you are: web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>
</web-app>

I can't find sun-jaxws.xml however... Thanks a lot! Regards

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The way to access a WSDL is not really container specific, it's more WS-stack specific. The WS-stack in GlassFish is Metro (Metro = JAX-WS RI + WSIT). Did you install/deploy Metro or JAX-WS RI on Tomcat? See Metro on Tomcat 6.x or Running JAX-WS Samples with Tomcat 6.x (JAX-WS RI might be enough in your case) for the steps.

Update: You need to declare the WSServlet in the web.xml (see Deploying Metro endpoint):

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
  <listener>
    <listener-class>
    com.sun.xml.ws.transport.http.servlet.WSServletContextListener
    </listener-class>
  </listener>
  <servlet>
    <servlet-name>WebServicePort</servlet-name>
    <servlet-class>
    com.sun.xml.ws.transport.http.servlet.WSServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>WebServicePort</servlet-name>
    <url-pattern>/services/*</url-pattern>
  </servlet-mapping>
  <session-config>
    <session-timeout>60</session-timeout>
  </session-config>
</web-app>

And then in the sun-jaxws.xml (also packaged in WEB-INF), declare your Service Endpoint Interface (SEI):

<?xml version="1.0" encoding="UTF-8"?>
<endpoints xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime" version="2.0">
  <endpoint
  name="MyHello"
  implementation="hello.HelloImpl"
  url-pattern="/hello"
  />
</endpoints>

And you access the WSDL at:

http://localhost:8080/<mycontext>/services/hello?wsdl
           A               B         C       D
  • A is the host and port of the servlet container.
  • B is the name of the war file.
  • C comes from the url-pattern element in the web.xml file.
  • D comes from the ending stem of the url-pattern attribute in the sun-jaxws.xml file.

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...