我试图在android应用程序中通过ksoap2开发连接mysql数据库。所以我提到了dis站点：
http://codeoncloud.blogspot.in/2012/03/android-mysql-client.html
我在我的应用程序中反复收到“02-28 06:16:40.267: DEBUG/SntpClient(66): request time failed: java.net.SocketException: Address family not supported by protocol”。
Dis是我的android编码：
package com.retailer.client;
import android.app.Activity;
import android.os.Bundle;
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.widget.TextView;
public class RetailerActivity extends Activity {
private static final String SOAP_ACTION = "retailer.com";
private static final String METHOD_NAME = "customerData";
private static final String NAMESPACE = "retailer.com";
private static final String URL = "http://192.168.1.249:8085/Retailer/services/RetailerWS?wsdl";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.setOutputSoapObject(request);
HttpTransportSE ht = new HttpTransportSE(URL);
try {
ht.call(SOAP_ACTION, envelope);
SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
SoapPrimitive s = response;
String str = s.toString();
String resultArr[] = str.split("&");//Result string will split & store in an array
TextView tv = new TextView(this);
for(int i = 0; i<resultArr.length;i++){
tv.append(resultArr[i]+"\n\n");
}
setContentView(tv);
} catch (Exception e) {
e.printStackTrace();
}
}
}
还为我的清单文件添加了 Internet 权限。
那为什么空白屏幕只显示在我的模拟器上..当我放置站点所有者 WSDL 链接时意味着它对我有用......但是我输入我的 url 意味着它对我不起作用..但是站点所有者 WSDL 文件和我的 WSDL文件都相同而已..我检查了这两个文件。然后y dis error dis来了。请指导我....