無法從JSONObject檢索值

Question

我通過PHP爲這個android java類（登錄屏幕）傳遞值，但它給了我一個JSONException錯誤，我無法從下面的代碼識別。 有人可以告訴我如何解決善意？先謝謝你！

LoginActivity.java

@Override 
public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.login); 

    // Importing all assets like buttons, text fields 
    inputUid = (EditText) findViewById(R.id.loginUid); 
    inputPassword = (EditText) findViewById(R.id.loginPassword); 
    btnLogin = (Button) findViewById(R.id.btnLogin); 

    btnLinkToRegister = (Button) findViewById(R.id.btnLinkToRegisterScreen); 

    loginErrorMsg = (TextView) findViewById(R.id.login_error); 

    // Login button Click Event 
    btnLogin.setOnClickListener(new View.OnClickListener() { 

     public void onClick(View view) { 
      String id = inputUid.getText().toString(); 
      String pswd = inputPassword.getText().toString(); 
      UserFunctions userFunction = new UserFunctions(); 
      Log.d("Button", "Login"); 
      JSONObject json = userFunction.loginUser(id, pswd); 

      // check for login response 
      try { 
       if (json.getString(KEY_SUCCESS) != null) { 
        loginErrorMsg.setText(""); 
        String res = json.getString(KEY_SUCCESS); 
        if(Integer.parseInt(res) == 1){ 
         // user successfully logged in 

         // Store user details in SQLite Database 
         DatabaseHandler db = new DatabaseHandler(getApplicationContext()); 
         JSONObject json_user = json.getJSONObject("user"); 

         // Clear all previous data in database 
         userFunction.logoutUser(getApplicationContext()); 
         db.addUser(json.getString(KEY_ID), json_user.getString(KEY_NAME), json_user.getString(KEY_EMAIL), json_user.getString(KEY_CREATED_AT));      

         // Launch Dashboard Screen 
         Intent dashboard = new Intent(getApplicationContext(), DashboardActivity.class); 

         // Close all views before launching Dashboard 
         dashboard.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP); 
         startActivity(dashboard); 

         // Close Login Screen 
         finish(); 
        }else{ 
         // Error in login 
         loginErrorMsg.setText("Incorrect username/password"); 
        } 
       } 
      } catch (JSONException e) { 
       e.printStackTrace(); 
      } 
     } 
    }); 

Answer 1

db.addUser(json.getString(KEY_ID), json_user.getString(KEY_NAME), json_user.getString(KEY_EMAIL), json_user.getString(KEY_CREATED_AT)); 

變化

db.addUser(json_user.getString(KEY_ID), json_user.getString(KEY_NAME), json_user.getString(KEY_EMAIL), json_user.getString(KEY_CREATED_AT));