我是新的web服務問題。我想通過用戶名和密碼連接到網絡服務器。我發現了一些解決方案,如何連接並獲得響應,但經過一些試驗後,我失敗了。我正在開始下面顯示的一些錯誤。這是我的代碼,你可以幫我嗎?通過使用用戶名和密碼連接到webservice
package com.isoft.uploader;
import java.util.ArrayList;
import org.json.JSONArray;
import org.json.JSONObject;
import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransport;
import android.app.Activity;
import android.content.Intent;
import android.database.Cursor;
import android.net.Uri;
import android.os.Bundle;
import android.provider.MediaStore;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
public class UploaderActivity extends Activity
{
//ArrayList<Response> WebData= new ArrayList<Response>();
public static final int SELECT_VIDEO=1;
public static final String TAG="UploadActivity";
String path="";
final String NAMESPACE = "http://tempuri.org/";
final String SERVICEURL = "http://192.168.10.177/androidws/isandroidws.asmx";
final String METHOD_NAME1="OzelVeriAlanlariniGetir";
final String METHOD_NAME="KullaniciGiris";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
Button enter=(Button)findViewById(R.id.Enter);
final EditText username=(EditText)findViewById(R.id.username);
final EditText password=(EditText)findViewById(R.id.password);
enter.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View arg0)
{
// TODO Auto-generated method stub
//request code for Webservice
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
//sending the username to the webservice
request.addProperty("kullaniciAdi",username.getText().toString());
//sending the password to the webservice
request.addProperty("password",password.getText().toString());
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
//Putting the request in an envelope
envelope.setOutputSoapObject(request);
HttpTransport transport = new HttpTransport(SERVICEURL);
try
{
transport.call("http://tempuri.org/"+METHOD_NAME, envelope);
//getting the response from the webservice
Boolean response = (Boolean) envelope.getResponse();
if(response==true)
{
openGaleryVideo();
}//end of if statement
else setContentView(R.layout.main);
}//end of try
catch(Exception exception)
{
exception.printStackTrace();
}//end of catch
}//end of onClick method
});//end of OnclickListener method
}//end of onCreate method
public void openGaleryVideo()
{
Intent intent=new Intent();
intent.setType("video/*");
intent.setAction(Intent.ACTION_GET_CONTENT);
startActivityForResult(Intent.createChooser(intent, "Select Video"),SELECT_VIDEO);
}//end of openGaleryVideo method
public String getPath(Uri uri)
{
String[] projection = { MediaStore.Video.Media.DATA};
Cursor cursor = managedQuery(uri, projection, null, null, null);
int column_index = cursor.getColumnIndexOrThrow(MediaStore.Video.Media.DATA);
cursor.moveToFirst();
return cursor.getString(column_index);
}//end of getPath method
}//end of main
你已經宣佈在AndroidManifest.xml所有的活動?另外,請檢查你的'UploaderActivity.java' - 第63行。看起來在那裏的'onClick'方法有問題。 – pixelscreen 2012-07-12 11:51:15
這是因爲它onClick,他試圖連接到web服務! – silentw 2012-07-12 11:52:33
是的我沒有任何問題,我認爲這是關於jar文件,但我不知道如何解決它。 – answer88 2012-07-12 11:54:15