Android的调用WCF服务 [英] Android call WCF service

问题描述

我有一个叫 WCF Web服务从 Android应用程序的一个问题。如果我叫下面的网址，http://10.0.2.2:80/WebService/WebServiceImpl.svc我得到的回应200 OK ，但是当我尝试调用服务中的函数http://10.0.2.2:80/WebService/WebServiceImpl.svc/Test我得到回应400错误的请求。

有人能帮助？

 命名空间的WebService
{
    公共类WebServiceImpl：IWebServiceImpl
    {
        ＃区域IRestServiceImpl成员
        公共字符串测试（）
        {
            返回检验合格;
        }        #endregion
    }
}WebService的命名空间
{
    [服务合同]
    公共接口IWebServiceImpl
    {
     [OperationContract的]
        [WebInvoke（
            方法=GET）]
        字符串Test（）;
    }
}
 

Android的活动：

 公共无效的onClick（视图v）
    {
        // TODO自动生成方法存根
        HttpClient的客户端=新DefaultHttpClient（）;
        字符串SERVER_HOST =10.0.2.2;
        INT SERVER_PORT = 80;
        字符串的URL =htt​​p://10.0.2.2:80/WebService/WebServiceImpl.svc/Test;
        HttpHost目标=新HttpHost（SERVER_HOST，SERVER_PORT，HTTP）;
        HTTPGET请求=新HTTPGET（URL）;
        尝试
        {
            HTT presponse响应= client.execute（目标，请求）;
            HttpEntity实体= response.getEntity（）;
            的MessageBox（response.getStatusLine（）的toString（））;
        }
        赶上（例外五）
        {
            的MessageBox（excepton）;
            的MessageBox（e.toString（））;
        }
    }    公共无效的MessageBox（字符串消息）{
        Toast.makeText（这一点，消息，Toast.LENGTH_LONG）.show（）;
    }
 

解决方案

我解决了使用SoapObject

code的一部分：

 公共静态最后弦乐NAMESPACE =htt​​p://tempuri.org/;
公共静态最终字符串URL =htt​​p://10.0.2.2:80/MyFirstPublishedWebService/WebServiceImpl.svc?wsdl;
公共静态最后弦乐SOAP_ACTION =htt​​p://tempuri.org/IWebServiceImpl/Login;
公共静态最后弦乐METHOD_NAME =登陆;SoapObject要求=新SoapObject空间（namespace，METHOD_NAME）;
SoapSerializationEnvelope信封=
    新SoapSerializationEnvelope（SoapEnvelope.VER11）;信封.dotNet = TRUE;envelope.setOutputSoapObject（请求）;
HttpTransportSE androidHttpTransport =新HttpTransportSE（URL）;尝试{
    androidHttpTransport.call（SOAP_ACTION，信封）;
}
赶上（例外五）{}
 

亚历克斯，感谢您的回答！

I have a problem with calling WCF web service from Android application. If I call the following URL, "http://10.0.2.2:80/WebService/WebServiceImpl.svc" I get the response "200 OK", but when I try to call the function within the service "http://10.0.2.2:80/WebService/WebServiceImpl.svc/Test" I get a response "400 Bad request".

Can someone help?

namespace WebService
{
    public class WebServiceImpl : IWebServiceImpl
    {
        #region IRestServiceImpl Members
        public string Test()
        {
            return "Test pass";
        }

        #endregion
    }
}

namespace WebService
{
    [ServiceContract]
    public interface IWebServiceImpl
    {
     [OperationContract]
        [WebInvoke(
            Method = "GET")]
        string Test();
    }
}

Android Activity:

    public void onClick(View v)
    {
        // TODO Auto-generated method stub
        HttpClient  client = new DefaultHttpClient();
        String SERVER_HOST="10.0.2.2";
        int SERVER_PORT = 80;
        String URL = "http://10.0.2.2:80/WebService/WebServiceImpl.svc/Test";
        HttpHost target = new HttpHost(SERVER_HOST, SERVER_PORT, "http");
        HttpGet request = new HttpGet(URL);
        try
        {
            HttpResponse response = client.execute(target,request);
            HttpEntity entity = response.getEntity();
            MessageBox(response.getStatusLine().toString());
        }
        catch(Exception e)
        {
            MessageBox("excepton");
            MessageBox(e.toString());
        }
    }

    public void MessageBox(String message){
        Toast.makeText(this,message,Toast.LENGTH_LONG).show();
    }

解决方案

I solve that using "SoapObject"

part of code:

public static final String NAMESPACE = "http://tempuri.org/";
public static final String URL = "http://10.0.2.2:80/MyFirstPublishedWebService/WebServiceImpl.svc?wsdl";  
public static final String SOAP_ACTION = "http://tempuri.org/IWebServiceImpl/Login";
public static final String METHOD_NAME = "Login";

SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
SoapSerializationEnvelope envelope = 
    new SoapSerializationEnvelope(SoapEnvelope.VER11); 

envelope .dotNet = true;

envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);

try {
    androidHttpTransport.call(SOAP_ACTION, envelope);
}
catch (Exception e) {

}

Alex, Thanks for answer!

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