2015-08-14 71 views
3

我試圖爲應用程序創建一個登錄。但是我有一個問題。方法getText()必須從UI線程調用(Android Studio)

這是我的代碼:

package com.forgetmenot.loginregister; 


import java.util.ArrayList; 
import java.util.List; 

import org.apache.http.NameValuePair; 
import org.apache.http.message.BasicNameValuePair; 
import org.json.JSONArray; 
import org.json.JSONException; 
import org.json.JSONObject; 

import android.app.Activity; 
import android.content.Intent; 
import android.os.AsyncTask; 
import android.os.Bundle; 
import android.util.Log; 
import android.view.Menu; 
import android.view.MenuItem; 
import android.view.View; 
import android.widget.Button; 
import android.widget.EditText; 

public class MainActivity extends Activity { 
    EditText uname, password; 
    Button submit; 
    // Creating JSON Parser object 
    JSONParser jParser = new JSONParser(); 
    private static final String TAG = "Login"; 

    JSONObject json; 
    private static String url_login = "http://localhost:8080/ForgetMeNotApplication/Login"; 
    //JSONArray incoming_msg = null; 
    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.activity_main); 
     findViewsById(); 
     submit.setOnClickListener(new View.OnClickListener() { 

      @Override 
      public void onClick(View arg0) { 
       // execute method invokes doInBackground() where we open a Http URL connection using the given Servlet URL 
       //and get output response from InputStream and return it. 
       new Login().execute(); 

      } 
     }); 
    } 
    private void findViewsById() { 

     uname = (EditText) findViewById(R.id.txtUser); 
     password = (EditText) findViewById(R.id.txtPass); 
     submit = (Button) findViewById(R.id.login); 
    } 
    private class Login extends AsyncTask<String, String, String>{ 

     @Override 
     protected String doInBackground(String... args) { 
      // Getting username and password from user input 

      String username = uname.getText().toString(); 
      String pass = password.getText().toString(); 

      List<NameValuePair> params = new ArrayList<NameValuePair>(); 
      params.add(new BasicNameValuePair("u",username)); 
      params.add(new BasicNameValuePair("p",pass)); 
      json = jParser.makeHttpRequest(url_login, "GET", params); 
      String s=null; 

      try { 
       s= json.getString("info"); 
       Log.d("Msg", json.getString("info")); 
       if(s.equals("success")){ 
        Intent login = new Intent(getApplicationContext(), home.class); 
        login.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP); 
        startActivity(login); 
        finish(); 
       } 
      } catch (JSONException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

      return null; 
     } 

    } 

    @Override 
    public boolean onCreateOptionsMenu(Menu menu) { 
     // Inflate the menu; this adds items to the action bar if it is present. 
     getMenuInflater().inflate(R.menu.menu_main, menu); 
     return true; 
    } 

    @Override 
    public boolean onOptionsItemSelected(MenuItem item) { 
     // Handle action bar item clicks here. The action bar will 
     // automatically handle clicks on the Home/Up button, so long 
     // as you specify a parent activity in AndroidManifest.xml. 
     int id = item.getItemId(); 
     if (id == R.id.action_settings) { 
      return true; 
     } 
     return super.onOptionsItemSelected(item); 
    } 
} 

的Android工作室說,法getText()必須從UI線程在istructions被稱爲:uname.getText().toString();password.getText().toString();可能的解決方案?

回答

12

嘗試通過execute(param1, param1, ..., paramN)方法傳遞您對Login AsyncTask值:

submit.setOnClickListener(new View.OnClickListener() { 

      @Override 
      public void onClick(View arg0) { 

       String username = uname.getText().toString(); 
       String pass = password.getText().toString(); 
       new Login().execute(username, pass); 

      } 
     }); 
    } 
    private void findViewsById() { 

     uname = (EditText) findViewById(R.id.txtUser); 
     password = (EditText) findViewById(R.id.txtPass); 
     submit = (Button) findViewById(R.id.login); 
    } 
    private class Login extends AsyncTask<String, String, String>{ 

     @Override 
     protected String doInBackground(String... args) { 
      // Getting username and password from user input 

      String username = args[0]; 
      String pass = args[1]; 
2

使usernamepasslogin類變量和覆蓋onPreExcecute()和做到這一點:

@Override 
    protected void onPreExecute() { 

     username = uname.getText().toString(); 
     pass = password.getText().toString(); 

    } 
0

拋出異常,因爲doInBackground()是從後臺線程調用。我會將兩個字符串參數添加到Login類的構造函數中。

我想建議您在Android上獲得更多關於AsyncTasks和Threading的知識。例如,官方文檔中有一個頁面:http://developer.android.com/guide/components/processes-and-threads.html

您也可以查看本課程:https://www.udacity.com/course/developing-android-apps--ud853。你可以學習很多Android框架的基礎知識。

1

你從後臺線程訪問UI線程在這裏:

String username = uname.getText().toString(); 
String pass = password.getText().toString(); 

你想要做的只是通過用戶名/密碼字符串到您的後臺任務的構造函數,或者你可以直接轉給什麼執行方法。我更喜歡將它們定義到構造函數中,如果它們將被要求的話(就像你的那樣)。

定義您LoginTask像

String uname; 
String password; 
public Login(String username, String password({ 
    this.uname = username; 
    this.password = password; 
} 

然後在doInBackground()使用成員來代替。

List<NameValuePair> params = new ArrayList<NameValuePair>(); 
params.add(new BasicNameValuePair("u",this.username)); 
params.add(new BasicNameValuePair("p",this.pass)); 
json = jParser.makeHttpRequest(url_login, "GET", params); 

編輯 - 這樣的話你的新登錄()execute()調用看起來更像是這個

new Login(uname.getText().toString(), password.getText().toString()).execute(); 
1

除非事情已經改變了,我是不知道的,那不應該。一個問題。用戶界面元素不能從後臺更新,但訪問他們的獲取者從來都不是問題。

無論如何,你可以通過向AsyncTask添加一個構造函數來解決這個問題,這需要兩個String然後在創建任務時發送它們。

private class Login extends AsyncTask<String, String, String>{ 

    // member variables of the task class 
    String uName, pwd 
    public Login(String userName, String password) { 
     uName = userName; 
     pwd = password; 
    } 

    @Override 
    protected String doInBackground(String... args) {...} 

,並通過他們在onClick()

@Override 
public void onClick(View arg0) { 
    // execute method invokes doInBackground() where we open a Http URL connection using the given Servlet URL 
    //and get output response from InputStream and return it. 

    // pass them here 
    new Login(uname.getText().toString(), password.getText().toString()).execute(); 
} 
+0

在我的API 19設備,我發現'的getText() '似乎工作,但其他方法,如'TextureView'的'getBitmap()',崩潰。 – VinceFior

+0

我認爲'getText'可以在工作線程中訪問,它只是ide提示,而不是運行時異常 – Ninja

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