Question

在我的onCreate函数中添加了ip部分检查后，我的移动应用程序停止工作，使注册用户绕过登录屏幕。这可能与变量声明有关，但我仍然不确定发生了什么。

private EditText etUsername;
private EditText etPassword;
private EditText etIpAddress;
private Button btnLogin;

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    String username = etUsername.toString();
    String password = etPassword.toString();
    String ipAddress = etIpAddress.toString();

    SharedPreferences sharedPreferences = PreferenceManager.getDefaultSharedPreferences(LoginActivity.this);
    if (sharedPreferences.contains("ip")) {
        performLogin(username, password, sharedPreferences.getString("ip", ipAddress));
    }

    // declaring variebles
    etUsername = (EditText)findViewById(R.id.etUsername);
    etPassword= (EditText)findViewById(R.id.etPassword);
    btnLogin = (Button)findViewById(R.id.btnLogin);
    etIpAddress = (EditText) findViewById(R.id.etIpAddress);

    // setting up things for login button
    btnLogin.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {

            String ipAddress = etIpAddress.getText().toString();

            SharedPreferences sharedPreferences = PreferenceManager.getDefaultSharedPreferences(LoginActivity.this);

            sharedPreferences.edit()
                    .putString("ip", ipAddress)
                    .apply();

            String username = etUsername.getText().toString().trim();
            String password = etPassword.getText().toString().trim();

            performLogin(username, password, ipAddress);
        }
    });
}

private void performLogin(String username, String password, String ipAddress) {
    try {
        Device.login(username, password, ipAddress, this);
    } catch (JSONException e) {
        onLoginFailure(e);
    }
}

logcat的：

06-06 16:40:11.408 4000-4000/com.itemlocator.findit E/AndroidRuntime: FATAL EXCEPTION: main
                                                                      Process: com.itemlocator.findit, PID: 4000
                                                                      java.lang.RuntimeException: Unable to start activity ComponentInfo{com.itemlocator.findit/com.itemlocator.findit.LoginActivity}: java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String android.widget.EditText.toString()' on a null object reference
                                                                          at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2416)
                                                                          at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2476)
                                                                          at android.app.ActivityThread.-wrap11(ActivityThread.java)
                                                                          at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1344)
                                                                          at android.os.Handler.dispatchMessage(Handler.java:102)
                                                                          at android.os.Looper.loop(Looper.java:148)
                                                                          at android.app.ActivityThread.main(ActivityThread.java:5417)
                                                                          at java.lang.reflect.Method.invoke(Native Method)
                                                                          at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:726)
                                                                          at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:616)
                                                                       Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String android.widget.EditText.toString()' on a null object reference
                                                                          at com.itemlocator.findit.LoginActivity.onCreate(LoginActivity.java:32)
                                                                          at android.app.Activity.performCreate(Activity.java:6237)
                                                                          at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1107)
                                                                          at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2369)
                                                                          at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2476) 
                                                                          at android.app.ActivityThread.-wrap11(ActivityThread.java) 
                                                                          at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1344) 
                                                                          at android.os.Handler.dispatchMessage(Handler.java:102) 
                                                                          at android.os.Looper.loop(Looper.java:148) 
                                                                          at android.app.ActivityThread.main(ActivityThread.java:5417) 
                                                                          at java.lang.reflect.Method.invoke(Native Method) 
                                                                          at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:726) 
                                                                          at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:616)

Answer 1

String username = etUsername.toString();
String password = etPassword.toString();
String ipAddress = etIpAddress.toString();

将生成NullPointerExceptions，因为您的EditTexts尚未初始化。你必须先打电话：

etUsername = (EditText)findViewById(R.id.etUsername);
etPassword= (EditText)findViewById(R.id.etPassword);
btnLogin = (Button)findViewById(R.id.btnLogin);
etIpAddress = (EditText) findViewById(R.id.etIpAddress);

我认为你的意思是

String username = etUsername.getText().toString();

Answer 2

在登录屏幕中，在login_WS调用之后输入此代码

`SharedPreferences pref = getActivity（）。getSharedPreferences（MyPre，Context.MODE_PRIVATE）; 编辑器编辑器= pref.edit（）; editor.putString（PARAMS.KEY_PROFILE_DATA，jsonObject.get（＆＃34; data＆＃34;）。toString（））; editor.putString（PARAMS.KEY_USER_ID，ConstantData.USER_ID）; editor.putBoolean（PARAMS.KEY_STAY_LOGGED_IN，true）; editor.commit（）;

在启动画面中输入此代码

`private Runnable mRunnable = new Runnable（）{

    @Override
    public void run() {

        pref = getSharedPreferences(MyPre, Context.MODE_PRIVATE);

        //Log.i("splash","Boolen value 1" + pref.getBoolean(PARAMS.KEY_STAY_LOGGED_IN,false));

         is = Boolean.parseBoolean(String.valueOf((pref.getBoolean(PARAMS.KEY_STAY_LOGGED_IN, false))));

        Log.i("splash","Boolen value 2" + is);

        if (is) {

            Intent mIntent = new Intent(SplashScreenActivity.this, MainActivity.class);
            startActivity(mIntent);
            // overridePendingTransition(R.anim.enter, R.anim.no_anim);
            finish();

        } else {

            Intent mIntent = new Intent(SplashScreenActivity.this, LoginActivity.class);
            startActivity(mIntent);
            overridePendingTransition(R.anim.enter, R.anim.no_anim);
            finish();

        }`

希望这对你有用

Android Studio共享首选项应用程序停止工作

2 个答案: