部署web Service到tomcat

Posted 2020-06-15

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打开jdeveloper 12c,然后新建一个java项目,点击java,生成web services.

package simple;

import javax.jws.WebMethod;
import javax.jws.WebParam;
import javax.jws.WebService;

@WebService
public class HelloWorld {
public HelloWorld() {
super();
}

@WebMethod
public String sayHelloWorld (@WebParam(name = "arg0") String username) {
return "Hello "+username;
}
}

在WEB-INF下添加sun-jaxws.xml,内容如下

<?xml version="1.0" encoding="GBK" ?>
<endpoints
xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime"
version="2.0">
<endpoint
name="HelloWorld"
implementation="simple.HelloWorld"
url-pattern="/HelloWorldPort"/>
</endpoints>

修改web.xml,内容如下:

<?xml version = ‘1.0‘ encoding = ‘GBK‘?>
<web-app 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_3_0.xsd"
version="3.0">
<listener>
<listener-class>
com.sun.xml.ws.transport.http.servlet.WSServletContextListener
</listener-class>
</listener>
<servlet>
<servlet-name>HelloWorldPort</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>HelloWorldPort</servlet-name>
<url-pattern>/HelloWorldPort</url-pattern>
</servlet-mapping>
</web-app>