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我想在 Android 项目上连接到我的 WCF WebService。我按照教程

http://javatutorialspoint.blogspot.com/2012/02/android-web-service-access-using-ksoap2.html

但我收到 415 错误。也许还有其他简单的连接方式？或者我应该修复什么？

04-16 10:57:10.675: E/WS(4858): java.io.IOException: HTTP 请求失败，HTTP 状态：415

这是我的网络服务的网址（它是本地的）http://hq.fs.com.pl:8090/fs/Integrator

活动

package com.example.fs;

import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapPrimitive;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;

import android.util.Log;
import android.widget.TextView;
import android.app.Activity;
import android.os.Bundle;

public class AndroidWSClientActivity extends Activity {

    //private static final String SOAP_ACTION = "http://ws.android.com/sayHello";
    //private static final String METHOD_NAME = "sayHello";
    //private static final String NAMESPACE = "http://ws.android.com/";
    //private static final String URL = "http://175.157.229.119:8080/AndroidWSTest/services/PrintMsg?wsdl";

    // WS
    private static final String SOAP_ACTION = "http://tempuri.org/IntegratorMethods/GetAddonsTypes";
    private static final String METHOD_NAME = "GetAddonsTypes";
    private static final String NAMESPACE = "http://tempuri.org/";
    private static final String URL = "http://hq.fs.com:8090/fs/Integrator?wsdl";

    TextView textView;
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_wsclient_page);

    textView = (TextView) findViewById(R.id.textView2);


    Thread networkThread = new Thread() {
        @Override
        public void run() {
          try {
             SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);         
             SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
             envelope.setOutputSoapObject(request);

             HttpTransportSE ht = new HttpTransportSE(URL);
             ht.call(SOAP_ACTION, envelope);
             final  SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
             final String str = response.toString();

             runOnUiThread (new Runnable(){ 
         public void run() {
             textView.setText(str);
               }
           });
          }
         catch (Exception e) {
            Log.e("WS", e.toString());
         }
        }
      };
      networkThread.start();
      }
}

更新

我用 chrome://soaclient/content/webservices.xul 检查了我与 WebService 的连接，一切看起来都很好。我有操作：GetAddonsTypes 和状态：200 OK。

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3 回答 3

3

我是一名中国开发人员，所以我的英语很差。我遇到了同样的问题，通过将 SoapEnvelope.VER12 更改为 SoapEnvelope.VER10 解决了这个问题。问题解决了，但是不知道为什么。我的 wsdl 文件是 xmlns:soap12="http://schemas.xmlsoap.org/wsdl/soap12/" 所以版本应该是 12，而不是 10。

于 2014-07-30T05:47:44.253 回答
1

我有同样的问题，通过在 WCF 服务的端点配置上将 WsHttpBinding 更改为 basicHttpBinding 来解决它

于 2013-05-31T21:05:54.797 回答
0

这可能会帮助你点Request赞

private static final String URL =
"http://give Here Server Ipaddress/AndroidWSTest/services/PrintMsg?wsl";

意味着提供当前更新ipaddress 也确保您的通过

(NAMESPACE, METHOD_NAME)

是正确的并且匹配webServie Method name并且Namespace还检查Intenet Conncetion和webservice工作正常Server的一面或不是

于 2013-04-16T09:12:10.267 回答