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axis调用两种作派的.net WebService

发布时间: 2012-11-19 10:18:51 作者: rapoo

axis调用两种风格的.net WebService
最近做一个接口需要调用.net提供的WebService,调试过程很痛苦,总是出现Server was unable to read request的错误,并且.net服务端怎么都接不到传送的参数,接到的都是null。在网上查了半天资料也没有解决,最后下了个vs2008自己来写.net的服务学习了一下,才把问题搞定。


Rpc风格的.net服务端代码:

using System;using System.Collections;using System.ComponentModel;using System.Data;using System.Linq;using System.Web;using System.Web.Services;using System.Web.Services.Protocols;using System.Xml.Linq;namespace WebService1{            [WebService(Namespace = "http://tempuri.org/")]    public class Service1 : System.Web.Services.WebService    {        [WebMethod]        [SoapRpcMethod(Action = "http://tempuri.org/hello", RequestNamespace = "http://tempuri.org/", ResponseNamespace = "http://tempuri.org/")]        public string hello(String s)        {            return "Hello,"+s;        }    }}


Document风格的.net服务端代码:
namespace WebService1{            [WebService(Namespace = "http://tempuri.org/")]    public class Service1 : System.Web.Services.WebService    {        [WebMethod]                public string hello(String s)        {            return "Hello,"+s;        }    }}


java客户端代码:
String url="http://localhost:1117/Service1.asmx";String namespace = "http://tempuri.org/";String methodName = "hello";String soapActionURI = "http://tempuri.org/hello";Service service = new Service();Call call = (Call) service.createCall();call.setTargetEndpointAddress(new java.net.URL(url));call.setUseSOAPAction(true);//这个地方没设对就会出现Server was unable to read request的错误call.setSOAPActionURI(soapActionURI);call.setOperationName(new QName(namespace, methodName));/*这里如果设置成call.addParameter(new QName(namespace,"s"), XMLType.XSD_STRING,   ParameterMode.IN);就是调用document风格的.net服务端  如果设反了,.net服务端就接不到参数,接到的是null*/call.addParameter("s", XMLType.XSD_STRING, ParameterMode.IN);call.setReturnType(XMLType.XSD_STRING);String ret = (String) call.invoke(new Object[] { "kusix" });System.out.println("返回结果---> " + ret);



如果.net在提供Rpc服务时设置成
[SoapRpcMethod(Action = "",.....

的话,那java客户端连
call.setUseSOAPAction(true);call.setSOAPActionURI(soapActionURI);

这2句都可以去掉了

1 楼 mazhi21tt 2008-08-06 好!捧场~~~~ 2 楼 ritcher 2008-12-04 讲得很清晰,解决了问题,引用了:)

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