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我很難嘗試將AsyncTask應用到我現有的代碼。如何使用現有代碼的異步任務
如此虐待開始說我的代碼工作正常，直到我調升我的目標SDK，而我得到了以下錯誤消息：
android.os.NetworkOnMainThreadException at android.os.StrictMode$AndroidBlockGuardPolicy.onNetwork(StrictMode.java:1133)
經過一番研究，看起來這是因爲我試圖在主線程上運行網絡操作，這是一個很大的禁忌。解決方法是使用異步任務來運行網絡操作。
好的，到目前爲止對我來說是有意義的，現在我所要做的就是以某種方式在我的代碼中實現異步任務，並且在這裏解決我的問題。
基本上，我有一個登錄屏幕，導致成功登錄的主頁（有點像Facebook）。當您單擊登錄按鈕時，它會向我的服務器上的PHP文件發送一個HTTP請求，驗證登錄名/密碼併發迴響應。基於這種反應，它會讓你登錄（或者給你一個「無效登錄」響應）。
所以我很確定這是罪魁禍首。現在，我的問題是，我該如何在一個Asynch？我不是在尋找任何人寫我的代碼或任何東西，我只是尋找一些指導如何讓這個開始。我已經在這個幾天了，現在我在武器:(
下面是代碼將我登錄類，你可以在底部看到我的非同步：
public class AndroidLogin extends Activity implements OnClickListener {
Button ok,back,exit;
TextView result;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
// Login button clicked
ok = (Button)findViewById(R.id.btn_login);
ok.setOnClickListener(this);
result = (TextView)findViewById(R.id.tbl_result);
}
public void postLoginData() {
// Add user name and password
EditText uname = (EditText)findViewById(R.id.txt_username);
String username = uname.getText().toString();
EditText pword = (EditText)findViewById(R.id.txt_password);
String password = pword.getText().toString();
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
// login.php returns true if username and password match in db
HttpPost httppost = new HttpPost("http://www.alkouri.com/android/login.php?username=" + username + "&password=" + password );
try {
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("username", username));
nameValuePairs.add(new BasicNameValuePair("password", password));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
Log.w("SENCIDE", "Execute HTTP Post Request");
HttpResponse response = httpclient.execute(httppost);
String str = inputStreamToString(response.getEntity().getContent()).toString();
Log.w("SENCIDE", str);
if(str.toString().equalsIgnoreCase("true"))
{
Log.w("SENCIDE", "TRUE");
result.setText("Login Successful! Please Wait...");
}else
{
Log.w("SENCIDE", "FALSE");
result.setText(str);
}
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
private StringBuilder inputStreamToString(InputStream is) {
String line = "";
StringBuilder total = new StringBuilder();
// Wrap a BufferedReader around the InputStream
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
// Read response until the end
try {
while ((line = rd.readLine()) != null) {
total.append(line);
}
} catch (IOException e) {
e.printStackTrace();
}
// Return full string
return total;
}
//when register button is clicked
public void RegisterButton(View view) {
Intent myIntent = new Intent(AndroidLogin.this, Registration.class);
AndroidLogin.this.startActivity(myIntent);
}
protected void onPostExecute(Void v){
// turns the text in the textview "Tbl_result" into a text string called "tblresult"
TextView tblresult = (TextView) findViewById(R.id.tbl_result);
// If "tblresult" text string matches the string "Login Successful! Please Wait..." exactly, it will switch to next activity
if (tblresult.getText().toString().equals("Login Successful! Please Wait...")) {
Intent intent = new Intent();
//take text in the username/password text boxes and put them into an extra and push to next activity
EditText uname2 = (EditText)findViewById(R.id.txt_username);
String username2 = uname2.getText().toString();
EditText pword2 = (EditText)findViewById(R.id.txt_password);
String password2 = pword2.getText().toString();
intent.putExtra("username2", username2 + "&pword=" + password2);
startActivity(intent);
}
}
public void onClick(View view) {
class PostLogingDataTask extends AsyncTask<Void,Void,Void>
{
protected Void doInBackground (Void... t)
{
postLoginData();
return null;
}
}
new PostLogingDataTask().execute();
}
}
這裏是我得到的錯誤：
Caused by: android.view.ViewRootImpl$CalledFromWrongThreadException: Only the original thread that created a view hierarchy can touch its views.
沒有在你的代碼中沒有的AsyncTask。 – Raghunandan
我剛剛修改我的代碼，我把它的異步任務@Raghunandan – user3155326
從nnackground線程更新/訪問ui是不可能的。這就是爲什麼你會得到例外 – Raghunandan