如何在Android中连接Web服务并从Web服务获取方法的结果？

Question

我创建了一个Web服务（.asmx）。有一个Login方法。它返回一个字符串。

DBClass myDBClass= new DBClass ();
public string ResultUserName{ get; set; }
[WebMethod ]
public string Login(string UserName,string Password)
{
    DataTable tb = new DataTable();
    string sql = @"Select UserName from  UserTable where UserName='"+UserName+"' and Password='"+Password+"' ";
    tb = myDBClass.Get(sql, "login");
    if (tb.Rows.Count > 0)
    {
        ResultUserName= Convert.ToString(tb.Rows[0]["UserName"]);

    }

    else
    {
        ResultUserName= "The UserName Is Not Found";
    }

    return ResultUserName;
}

我正在使用Android的ksoap2库。

我为连接服务创建了一个类，并从服务中获取结果。

CallSoap.java

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

public class CallSoap {

public final String SOAP_ACTION = "http://tempuri.org/Login";
public  final String OPERATION_NAME = "Login";
public  final String WSDL_TARGET_NAMESPACE = "http://tempuri.org/";
public  final String SOAP_ADDRESS = "http://myserver.com:88/Sample.asmx";

public String Login(String UserName,String Password)

{
    Object response=null;
    SoapObject request = new SoapObject(WSDL_TARGET_NAMESPACE,OPERATION_NAME);
    PropertyInfo pi;
    PropertyInfo pi2;

    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    HttpTransportSE httpTransport = new HttpTransportSE(SOAP_ADDRESS);
    pi=new PropertyInfo();
    pi.setName("UserName");
    pi.setValue(UserName);
    pi.setType(PropertyInfo.STRING_CLASS);
    request.addProperty(pi);
    pi2=new PropertyInfo();
    pi2.setName("Password");
    pi2.setValue(Password);
    pi2.setType(PropertyInfo.STRING_CLASS);
    request.addProperty(pi2);

    try
    {
        httpTransport.call(SOAP_ACTION, envelope);
        response = envelope.getResponse();
    }
    catch (Exception exception)
    {
        response=exception.toString();
    }
    return response.toString();


 }
    }

activity_my.xml有1个Buttton，2个EditText和1个TextView

<Button
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:text="Log In"
    android:id="@+id/btnLogin"
    android:layout_alignParentTop="true"
    android:layout_centerHorizontal="true" />
<TextView
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:id="@+id/lblResult"
    android:text="Result"
    android:layout_marginTop="114dp"
    android:layout_alignParentTop="true"
    android:layout_centerHorizontal="true" />

<EditText
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:hint="UserName"
    android:id="@+id/txtUserName"
    android:layout_toStartOf="@+id/btnLogin"
    android:layout_centerVertical="true"
    android:layout_toLeftOf="@+id/btnLogin" />

<EditText
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:ems="10"
    android:password="true"
    android:hint="Password"
    android:id="@+id/txtPassword"
    android:layout_alignTop="@+id/txtUserNmae"
    android:layout_toEndOf="@+id/btnLogin"
    android:layout_toRightOf="@+id/btnLogin" />

MyActivity.java

   @Override
   protected void onCreate(Bundle savedInstanceState) {
   super.onCreate(savedInstanceState);
   setContentView(R.layout.activity_my);
   Button btnLogin= (Button) findViewById(R.id.btnLogin);
   btnCall.setOnClickListener(new View.OnClickListener() {

   @Override
   public void onClick(View view) {
   EditText txtUserName=(EditText)findViewById(R.id.txtUserName);
   EditText txtPassword=(EditText)findViewById(R.id.txtPassword);
   TextView lblResult=(TextView)findViewById(R.id.lblResult);
   String strUserName=txtUserName.getText().toString();
   String strPassword=txtPassword.getText().toString();

   if (android.os.Build.VERSION.SDK_INT > 9)
   {

    StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
        StrictMode.setThreadPolicy(policy);
    }

  if(TextUtils.isEmpty(strUserName))
  {

   txtUserName.setError("Enter a UserName");
  }

  else if(TextUtils.isEmpty(strPassword))
 {
  txtPassword.setError("Enter a Password");

 }

  else
  {
    CallSoap cs=new CallSoap();

      try

      {
          String Result= cs.Login(strUserName, strPassword).toString();

          if (Result==strUserName) {

              lblResult.setText("Welcome:" + " " + Result);

          }
          else

          {
              lblResult.setText("UserName is not found");
              txtPassword.setText("");

          }

      }
      catch (Exception e)
      {
          lblResult.setText(e.toString());

      }

  }


}

});


}

的AndroidManifest.xml

<uses-permission android:name="android.permission.INTERNET" />

我的数据库中有此用户用户：示例密码：12345

当我尝试从Android登录时，我得到：＆＃34;找不到用户名＆＃34;。它无法从服务中获得结果。

我的错误在哪里？我没找到。