axis2实现webservice服务和tomcat集成

古城微笑少年丶 2022-05-12 06:44 193阅读 0赞

# axis2实现webservice服务和tomcat集成 #

通过上一篇[axis2实现webservice服务][axis2_webservice]实现服务端。

## 创建web工程 ##

*  在idea中新建web工程，并把axis2需要的lib包放入到web-inf/lib中  
    \\axis2-1.7.8\\webapp\\WEB-INF\\lib\\  
    \\axis2-1.7.8\\lib\\  
    并在idea的工程中配置，把lib文件夹添加到工程的依赖包中。
 *  编辑web.xml，添加如下内容:

<servlet>
            <display-name>Apache-Axis Servlet</display-name>
            <servlet-name>AxisServlet</servlet-name>
            <servlet-class>org.apache.axis2.transport.http.AxisServlet</servlet-class>
            
            
            
            
            
            
            
            
            
            
            <load-on-startup>1</load-on-startup>
        </servlet>
    
    
        <servlet-mapping>
            <servlet-name>AxisServlet</servlet-name>
            <url-pattern>/servlet/AxisServlet</url-pattern>
        </servlet-mapping>
    
        <servlet-mapping>
            <servlet-name>AxisServlet</servlet-name>
            <url-pattern>*.jws</url-pattern>
        </servlet-mapping>
    
        <servlet-mapping>
            <servlet-name>AxisServlet</servlet-name>
            <url-pattern>/services/*</url-pattern>
        </servlet-mapping>
    
    
        <mime-mapping>
            <extension>inc</extension>
            <mime-type>text/plain</mime-type>
        </mime-mapping>

*  把MyServerImpl.aar拷贝到WEB-INF/services  
    目录结构如下:  
    ![这里写图片描述][70]

然后在idea中添加tomcat运行，启动tomcat就可以正确访问啦。  
在页面输入 [http://127.0.0.1:8080/services/MyServerImpl?wsdl][http_127.0.0.1_8080_services_MyServerImpl_wsdl]

如果客户端访问，需要修改uri路径为 [http://127.0.0.1:8080/services/MyServerImpl?wsdl][http_127.0.0.1_8080_services_MyServerImpl_wsdl]

[axis2_webservice]: https://blog.csdn.net/robin912/article/details/82712794
[70]: /images/20220512/9116728190f04856bbb9d3dee6fe4814.png
[http_127.0.0.1_8080_services_MyServerImpl_wsdl]: http://127.0.0.1:8080/services/MyServerImpl?wsdl