Http urlConnection ve Asyntask ile İstek Al

Question

Kullanıcıya iki EditText içeren bir arabirim istediğim kullanıcı adı ve parola paramları ile bir http isteği yapmaya çalışıyorum.

Aşağıdaki hatayı şu anda alıyorum: "DoInBackground()" yürütülürken bir hata oluştu. Yardım için teşekkürler.

Bu

açıktır: Tüm

<?xml version="1.0" encoding="utf-8"?> 
<manifest xmlns:android="http://schemas.android.com/apk/res/android" 
    package="com.example.leonelbaidal.moodlenots"> 

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

    <application 
     android:allowBackup="true" 
     android:icon="@mipmap/ic_launcher" 
     android:label="@string/app_name" 
     android:supportsRtl="true" 
     android:theme="@style/AppTheme"> 
     <activity 
      android:name=".MainActivity" 
      android:screenOrientation="portrait" > 
      <intent-filter> 
       <action android:name="android.intent.action.MAIN" /> 
       <category android:name="android.intent.category.LAUNCHER" /> 
      </intent-filter> 
     </activity><!-- ATTENTION: This was auto-generated to add Google Play services to your project for 
    App Indexing. See https://g.co/AppIndexing/AndroidStudio for more information. --> 
     <meta-data 
      android:name="com.google.android.gms.version" 
      android:value="@integer/google_play_services_version" /> 
    </application> 

</manifest> 

Ana Etkinlik:

public class MainActivity extends Activity { 

    HttpURLConnection urlConnection = null; 
    InputStream is; 
    private GoogleApiClient client; 

    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     final EditText password; 
     final EditText username; 
     Button login; 

     super.onCreate(savedInstanceState); 
     requestWindowFeature(Window.FEATURE_NO_TITLE); 
     setContentView(R.layout.activity_main); 

     username = (EditText) findViewById(R.id.txt_username); 
     password = (EditText) findViewById(R.id.txt_password); 
     login = (Button) findViewById(R.id.btn_loggin); 

     password.setTypeface(Typeface.DEFAULT); 
     password.setTransformationMethod(new PasswordTransformationMethod()); 

     final String httpPath = "http://10.0.23.34/android/login.php?u=" + username.getText().toString() + "&p=" + password.getText().toString(); 

     login.setOnClickListener(new View.OnClickListener() { 
      @Override 

      public void onClick(View v){ 
       new MyTask().execute(); 
      } 
     }); 

     client = new GoogleApiClient.Builder(this).addApi(AppIndex.API).build(); 
    } 

Ve görev sınıfı: Sonunda

private class MyTask extends AsyncTask<String, Void, Void>{ 

     String textResult; 

     @Override 
     protected Void doInBackground(String... httpPath){ 


      try { 
       URL url = new URL(httpPath[0]); 
       urlConnection = (HttpURLConnection) url.openConnection(); 
       urlConnection.setDoOutput(true); 

       BufferedReader in = new BufferedReader(new InputStreamReader(urlConnection.getInputStream())); 
       String stringBuffer; 
       String stringText = ""; 
       while ((stringBuffer = in.readLine()) != null) { 
        stringText += stringBuffer; 
       } 

       in.close(); 
       textResult = stringText; 

      } catch (MalformedURLException e) { 
       e.printStackTrace(); 
       textResult = e.toString(); 
      } catch (IOException e){ 
       e.printStackTrace(); 
       textResult = e.toString(); 
      } 
      return null; 
     } 

     @Override 
     protected void onPostExecute(Void result){ 
      Toast.makeText(getApplicationContext(), textResult, Toast.LENGTH_SHORT).show(); 
     } 
    } 

i App uygulamaya kodlu OtomatikOluşturuldu gelmiş İndeksleme API'sı.

Answer 1

Sen url giriş param

new MyTask().execute(httpPath); 

eksik