如何调用别人提供的webservice接口

传递对方的url,方法名，参数信息。不需要引用对方的代码。

测试跑通。

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package com.flyfox.client;

import org.apache.axis.client.Call; import org.apache.axis.client.Service; public class HelloWorldServiceClient { ??? public static void main(String[] args) throws Exception { ??????? Service service = new Service(); ??????? try { ??????????? Call call = (Call) service.createCall(); ??????????? call ??????????????????? .setTargetEndpointAddress("http://172.16.192.73:8080/np-fy-ajpc/services/ajpcTjpcService?wsdl"); ??????????? call.setOperationName("tjpcProcess"); ??????????? call.addParameter( ??????????????? "xmlParam",//设置要传递的参数 ??????????????? org.apache.axis.encoding.XMLType.XSD_STRING,??????????????? javax.xml.rpc.ParameterMode.IN); ??????????? call.setReturnType(org.apache.axis.encoding.XMLType.XSD_STRING); ??????????? //??????????? call.setTimeout(5000); ??????????? call.setUseSOAPAction(true); ??????????? call ??????????????????? .setSOAPActionURI("http://support.ajpc.fy.np.thunisoft.com/tjpcProcess"); ??????????? String result = (String) call ??????????????????? .invoke(new Object[] { "<QueryParam NAjlb='1' NSpcx='2' CBhAj='6A10F8068A8015310BA2887DEFFC8960' NSyfy='1'></QueryParam>" }); ??????????? System.out.println(result); ??????? } catch (Exception ex) { ??????????? ex.printStackTrace(); ??????? } ??? } }