使用jdk的jws实现 开发web services

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首先web.xml 必须要有对org.springframework.web.context.ContextLoaderListener支持

<listener>   <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

java 文件： 

import javax.jws.soap.SOAPBinding.Style;

@Service("soapService") 
@WebService(serviceName = "soap")
@SOAPBinding(style=Style.RPC)
public class SOAPImpl {
    
    @Autowired
    private DictionaryManager dictManger;
    
    @WebMethod
    public String datacatalog(String datacatalogcode) {
         List<FDatacatalog> datalog = dictManger.getUserCdctgs();        
        return datalog.get(0).getCatalogname();
    }

}

beans.xml(spring配置文件)

<beans xmlns="http://www.springframework.org/schema/beans"
 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p"
 xmlns:aop="http://www.springframework.org/schema/aop" xmlns:context="http://www.springframework.org/schema/context"
 xmlns:jee="http://www.springframework.org/schema/jee" xmlns:tx="http://www.springframework.org/schema/tx"
 xmlns:task="http://www.springframework.org/schema/task"
 xsi:schemaLocation="
   http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd
   http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
   http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd
   http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd
   http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
   http://www.springframework.org/schema/task http://www.springframework.org/schema/task/spring-task-3.0.xsd">

 <context:component-scan base-package="com.centit.dde.cxf" />

 <bean class="org.springframework.remoting.jaxws.SimpleJaxWsServiceExporter">
  <property name="baseAddress" value="http://localhost:8081/" />
 </bean>

</beans>