Java 在 Tomcat 上访问 wsdl
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accessing wsdl on Tomcat
提问by Krt_Malta
I have a web service and I was deploying it on GlassFish. I accessed its wsdl through http://localhost:10697/APIService/APIServiceService?wsdl.
我有一个 Web 服务，我将它部署在 GlassFish 上。我通过http://localhost:10697/APIService/APIServiceService?wsdl访问了它的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?wsdlbut I'm getting a 404 error. I tried various combinations but none seem to work.
现在我将 WAR 文件移植到 Tomcat 6.0.24 并部署它。但是，我正在尝试使用http://localhost:8080/APIService/APIServiceService?wsdl访问其wsdl，但出现 404 错误。我尝试了各种组合，但似乎都不起作用。
How can I access the wsdl file plz?
请问如何访问wsdl文件？
Thanks and regards,
感谢致敬，
Update:Here you are: web.xml:
更新：你在这里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.xmlhowever... Thanks a lot! Regards
sun-jaxws.xml但是我找不到...非常感谢！问候
采纳答案by Pascal Thivent
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.xor Running JAX-WS Samples with Tomcat 6.x(JAX-WS RI might be enough in your case) for the steps.
访问 WSDL 的方式实际上并不是特定于容器的，而是更特定于 WS 堆栈的。GlassFish 中的 WS 堆栈是 Metro（Metro = JAX-WS RI + WSIT）。您是否在 Tomcat 上安装/部署了 Metro 或 JAX-WS RI？有关步骤，请参阅Metro on Tomcat 6.x或Running JAX-WS Samples with Tomcat 6.x（在您的情况下 JAX-WS RI 可能就足够了）。
Update:You need to declare the WSServletin the web.xml(see Deploying Metro endpoint):
更新：您需要WSServlet在web.xml（请参阅部署 Metro 端点）中声明：
<?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):
然后在sun-jaxws.xml（也打包在 WEB-INF 中）中，声明您的服务端点接口 (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:
您可以在以下位置访问 WSDL：
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.
- A 是 servlet 容器的主机和端口。
- B 是war 文件的名称。
- C 来自 web.xml 文件中的 url-pattern 元素。
- D 来自 sun-jaxws.xml 文件中 url-pattern 属性的结尾词干。
