Spring整合CXF发布及调用WebService

给你一囗甜甜゛ 提交于 2020-01-02 15:23:47

这几天终于把webService搞定,下面给大家分享一下发布webService和调用webService的方法

  1. 添加jar包 (官方下载地址:http://cxf.apache.org/download.html 为了方便扩展我把所有的jar包都加了)
  2. 在web.xml中配置spirng以及CXF
    <?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">
      <display-name></display-name>    
      <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
      </welcome-file-list>
      <!-- spring -->
      <context-param>
          <param-name>contextConfigLocation</param-name>
          <param-value>classpath:applicationContext.xml</param-value>
      </context-param>
      
      <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
      </listener>
      
      
      <listener>
          <listener-class>org.springframework.web.util.IntrospectorCleanupListener</listener-class>
      </listener>
      
        <!-- CXFService -->
        <servlet>
            <servlet-name>CXFService</servlet-name>
            <servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
        </servlet>
        
        <servlet-mapping>
            <servlet-name>CXFService</servlet-name>
            <url-pattern>/*</url-pattern>
        </servlet-mapping>
    </web-app>

    3.写接口实现类   要发布webService的类及要被调用的接口

    package cn.com.lrl.service.impl;
    
    import javax.jws.WebParam;
    import javax.jws.WebResult;
    import javax.jws.WebService;
    
    import cn.com.lrl.service.JAXService;
    @WebService
    public class JAXServiceImpl implements JAXService {
    
        @Override
        public @WebResult(name="news")String sayHello(@WebParam(name="name") String name) {
            return name +"   hello!";
        }

    4.配置spirng配置文件,发布webService  jaxws:server 标签中的address 中的路径 就是访问webService的路径  实际路径是http://localhost:8080/ws/Users?wsdl(ws是项目名)

    <?xml version="1.0" encoding="UTF-8"?>
    
    <beans xmlns="http://www.springframework.org/schema/beans"
            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xmlns:jaxws="http://cxf.apache.org/jaxws"
            xsi:schemaLocation="
                http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
                http://cxf.apache.org/jaxws http://cxf.apache.org/schemas/jaxws.xsd"
                >
    
            <!-- 发布的webService类 -->    
            <bean id="userServiceBean" class="cn.com.lrl.service.impl.JAXServiceImpl"/>
            
            <!-- 注意下面的address,这里的address的名称就是访问的WebService的name -->
            <jaxws:server id="userService" serviceClass="cn.com.lrl.service.JAXService" address="/Users">
                <jaxws:serviceBean>
                    <!-- 要暴露的 bean 的引用 -->
                    <ref bean="userServiceBean"/>
                </jaxws:serviceBean>
                <!-- 添加请求日志 -->
                <jaxws:inInterceptors> 
                    <bean class="org.apache.cxf.interceptor.LoggingInInterceptor"/> 
                </jaxws:inInterceptors> 
                   <!-- 添加响应 日志 -->
                <jaxws:outInterceptors> 
                    <bean class="org.apache.cxf.interceptor.LoggingOutInterceptor"/> 
                   </jaxws:outInterceptors> 
            </jaxws:server>
    </beans>

    5.启动tomcat发布webService 访问你的webService  如果是这样就代表成功了6.通过wsimport解析wsdl文件

  

  7.将生成的代码复制到要调用webService的项目

  8.测试接口是否可以使用

public class Test {
    public static void main(String[] args) {
        JAXServiceService jaxServiceService =new JAXServiceService();
        JAXService jaxService=jaxServiceService.getJAXServicePort();
        //调用方法输出
        System.out.println(jaxService.sayHello("lrl"));
    }
}

  9.调用成功 控制台输出

  

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