如何在Tomcat7中部署WS

Question

我有一个带有类和JSP的Web应用程序，我使用mvn clean package -Dmaven.test.skip=true打包创建战争

我在应用程序中有这个类：

import javax.jws.WebMethod;
import javax.jws.WebService;
import javax.servlet.http.HttpSession;
import javax.xml.ws.WebServiceContext;
import javax.xml.ws.handler.MessageContext;

    @WebService(serviceName="IberiaWS")
    public class IberiaWS {
      @Resource
      private WebServiceContext wsContext;  
      public IberiaWS () {
      }
      private UserVO getSessionUserVO() {
        MessageContext mc = wsContext.getMessageContext();
        HttpSession session = ((javax.servlet.http.HttpServletRequest)mc.get(MessageContext.SERVLET_REQUEST)).getSession();
        return (UserVO)session.getAttribute("uservo");
      }
      private void setSessionUserVO(UserVO uservo) {
        MessageContext mc = wsContext.getMessageContext();
        HttpSession session = ((javax.servlet.http.HttpServletRequest)mc.get(MessageContext.SERVLET_REQUEST)).getSession();
        session.setAttribute("uservo", uservo);
      }
      @WebMethod
      public boolean login(String loginName, String loginPwd) throws Exception {
        this.setSessionUserVO(new UserDAO().findUser("_"+loginName, "__"+loginPwd));
        return isConnected();
      }
      @WebMethod
      public boolean isConnected() {
        return (this.getSessionUserVO()!=null);
      }
      @WebMethod
      public IberiaPerson getPerson(String id) {
        return new IberiaPerson();   
      }
      @WebMethod
      public IberiaPerson findPerson(String companyNr) {
        UserVO uservo = this.getSessionUserVO();
        IberiaPerson ret=null;
        PersonVO p= new PersonDAO().findByCompanyNr(uservo.getAdminCenterId(), uservo.getUserId(), "Iberia", companyNr);
        if (p!=null) {
          ret = new IberiaPerson();
          ret.setPersonId(p.getPersonId());
          ret.setCompanyName(p.getVehicleOwnerName());
          ret.setCategoryName(p.getCategoryName());
          ret.setCompanyNr(p.getCompanyNr());
          ret.setFirstName(p.getFirstName());
          ret.setLastName(p.getLastName());
          ret.setStatusId(p.getStatusId());
          ret.setGroupName(p.getGroupList());
          ret.setKeyCode(p.getKeyString());   
          ret.setComments(p.getLmComment());
        }
        return ret;   
      }
    }

我创建战争并将其部署在雄猫中。我想访问WS的WSDL，但我不知道该怎么做。显然我没有创建应用程序，我只是得到了源代码，但我不知道如何访问 WS。由于应用程序部署在上下文 iberiafleet 中，我尝试 http://localhost:8080/iberiafleet/IberiaWSPort?WSDL 但收到 HTTP 404 错误

这是该项目的网络.xml

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >
<web-app>
  <display-name>Archetype Created Web Application</display-name>
</web-app>

有没有办法使用 maven 构建、打包和部署 WS

？

我已经更改了网络.xml，但我也无法访问

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >
<web-app>
  <display-name>Archetype Created Web Application</display-name>
  <servlet>
    <servlet-name>
      IberiaWSPort
    </servlet-name>
    <servlet-class>
      com.iberia.webservice.IberiaWS
    </servlet-class>
    <load-on-startup>
      1
    </load-on-startup>
  </servlet>
 <servlet-mapping>
    <servlet-name>
      IberiaWSPort
    </servlet-name>
    <url-pattern>
      /IberiaWSPort
    </url-pattern>
  </servlet-mapping>
</web-app>

Answer 1

请按照以下示例代码进行操作

请参考链接

<?xml version="1.0" encoding="UTF-8"?><endpoints  xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime"  version="2.0">  
<endpoint name="AccountDetailsServiceEndPoint"    service ="{http://gognamunish.com/accounts}AccountDetailsService"      port="{http://gognamunish.com/accounts}AccountDetailsPort"      implementation="com.mg.ws.impl.AccountDetailsServiceImpl"      url-pattern="/details"      wsdl="WEB-INF/wsdl/accounts.wsdl"/></endpoints>