如何修復Android工作室中的登錄錯誤

Question

我試圖使用json進行登錄。我的php返回這個結果：

[{"name":"cynthia","password":"123456"},{"name":"John","password":"123456"}] 

這些是mySQL中的數據。

，我想利用這個結果在機器人工作室的比較，但我發現它不工作。它總是去到else語句。

public class LoginActivity extends AppCompatActivity { 

    public EditText username; 
    public EditText password; 
    private ArrayList<String> usernameList = new ArrayList<String>(); 
    private ArrayList<String> passwordList = new ArrayList<String>(); 

    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.activity_login); 
     username = (EditText) findViewById(R.id.username); 
     password = (EditText) findViewById(R.id.password); 
    } 

    public void Login2(View view) { 
     if (username.getText().length() == 0 || password.getText().length() == 0) { 
      openDialog(view); 
     } else { 
      new AsyncRetrieve().execute(); 
      if(usernameList.contains(username)&&passwordList.contains(password)){ 
       successLogin(); 
      }else{ 
       openDialog(view); 
      } 

     } 
    } 

    public void openDialog(View view) { 
     new AlertDialog.Builder(this) 
       .setTitle(R.string.login_dialog_title) 
       .setMessage(R.string.login_dialog_msg) 
       .show(); 
    } 

    public void successLogin() { 
     Intent intent = new Intent(); 
     intent.setClass(this, HomePageActivity.class); 
     startActivity(intent); 
    } 

    private class AsyncRetrieve extends AsyncTask<Void, Void, String> { 
     @Override 
     protected String doInBackground(Void... params) { 
      return getVideoList(); 
     } 

     private String getVideoList() { 
      try { 
       URL url = new URL("http://localhost/soften/login_check1.php"); 
       HttpURLConnection conn = (HttpURLConnection) url.openConnection(); 
       InputStream in = new BufferedInputStream(conn.getInputStream()); 
       BufferedReader reader = new BufferedReader(new InputStreamReader(in)); 
       String response = reader.readLine(); 

       //set the response from server as json array 
       JSONArray userArray = new JSONArray(response); 

       usernameList.clear(); 
       passwordList.clear(); 
       Log.d("username", "value = " + userArray.length()); 
       for (int i = 0; i < userArray.length(); i++) { 
        Log.d("i", "i = " + i); 
        //get json object from the json array 
        JSONObject json = userArray.getJSONObject(i); 
        Log.d("json", "" + json.length()); 
        //add object to the list for grid view 
        usernameList.add(json.getString("name")); 
        passwordList.add(json.getString("password")); 
       } 



       return null; 
      } catch (Exception ex) { 
       return ex.toString(); 
      } 
     } 
    } 
} 

Answer 1

你long2（）是埃羅米，是有

if(usernameList.contains(username)&&passwordList.contains(password)) 

你應該等於username.getText（）和password.getText（）。我的英語不是很好，但你應該能夠理解

Answer 2

的問題是在這條線

new AsyncRetrieve().execute(); 

當你執行的AsyncTask它開始一回地面線程，但立即進行下去，而不等待來自服務器的結果，並且當您比較用戶名和密碼時，這兩個列表都是空的。

private ArrayList<String> usernameList = new ArrayList<String>(); 
private ArrayList<String> passwordList = new ArrayList<String>(); 

解決方案：

首先解決方案是一個get（）方法添加到您的AsyncTask調用，比如

new AsyncRetrieve().execute().get(); 

，直到從服務器，然後收到的結果也將暫停你的AsyncTask做了你的比較。

另一解決方案是重寫onPostExecute（）中的AsyncTask和那裏進行比較。

private class AsyncRetrieve extends AsyncTask<Void, Void, String> { 
     @Override 
     protected String doInBackground(Void... params) { 
      return getVideoList(); 
     } 

     private String getVideoList() { 
      try { 
       URL url = new URL("http://localhost/soften/login_check1.php"); 
       HttpURLConnection conn = (HttpURLConnection) url.openConnection(); 
       InputStream in = new BufferedInputStream(conn.getInputStream()); 
       BufferedReader reader = new BufferedReader(new InputStreamReader(in)); 
       String response = reader.readLine(); 

       //set the response from server as json array 
       JSONArray userArray = new JSONArray(response); 

       usernameList.clear(); 
       passwordList.clear(); 
       Log.d("username", "value = " + userArray.length()); 
       for (int i = 0; i < userArray.length(); i++) { 
        Log.d("i", "i = " + i); 
        //get json object from the json array 
        JSONObject json = userArray.getJSONObject(i); 
        Log.d("json", "" + json.length()); 
        //add object to the list for grid view 
        usernameList.add(json.getString("name")); 
        passwordList.add(json.getString("password")); 
       } 



       return null; 
      } catch (Exception ex) { 
       return ex.toString(); 
      } 
     } 


     @Override 
     protected String onPostExecute(String result) { 
      if(usernameList.contains(username.getText())&&passwordList.contains(password.getText())){ 
       successLogin(); 
      }else{ 
       openDialog(view); 
      } 
     } 
    } 

取下Login2身份方法，這些行:)不要忘記添加的getText（）方法。

希望得到這個幫助。