Question

我正在尝试创建一个简单的登录页面，使用正确的用户名和密码将转到另一个活动（已经创建）此外，尝试计数器无法正常运行。对不起，如果说这个措辞不好，希望它有意义。

public class login extends Activity {

private EditText username;
private EditText password;
private Button login;
private TextView loginLocked;
private TextView attemptsLeft;
private TextView numberOfRemainingLoginAttemptsTV;
int numberOfRemainingLoginAttempts = 3;

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

public void Login(View view) {
    if (username.getText().toString().equals("admin") && password.getText().toString().equals("secret")) {
        Intent i = new Intent(login.this, MainActivity.class);
        startActivity(i);
        Toast.makeText(getApplicationContext(), "Welcome User",
                Toast.LENGTH_SHORT).show();

    } else {
        Toast.makeText(getApplicationContext(), "Wrong Credentials",
                Toast.LENGTH_SHORT).show();
        numberOfRemainingLoginAttempts--;
        attemptsLeft.setVisibility(View.VISIBLE);
        numberOfRemainingLoginAttemptsTV.setVisibility(View.VISIBLE);
        numberOfRemainingLoginAttemptsTV.setText(Integer.toString(numberOfRemainingLoginAttempts));

        if (numberOfRemainingLoginAttempts == 0) {
            login.setEnabled(false);
            loginLocked.setVisibility(View.VISIBLE);
            loginLocked.setBackgroundColor(Color.RED);
            loginLocked.setText("Please Try Again Later");
        }
    }
}

private void setupVariables() {
    username = (EditText) findViewById(R.id.username);
    password = (EditText) findViewById(R.id.password);
    login = (Button) findViewById(R.id.button);
    loginLocked = (TextView) findViewById(R.id.loginLocked);
    attemptsLeft = (TextView) findViewById(R.id.attemptsLeft);
    numberOfRemainingLoginAttemptsTV = (TextView) findViewById(R.id.numberOfRemainingLoginAttemptsTV);
    numberOfRemainingLoginAttemptsTV.setText(Integer.toString(numberOfRemainingLoginAttempts));
}
}

Answer 1

当你使用startActivity()时，你必须在上下文中使用它（并不总是。仅在某些条件下，但我认为在上下文中声明它是一个好习惯）。因此，将startActivity(i)更改为login.this.startActivity(i);

希望这应该让它运行。希望能帮助到你！

编辑：也请发布您的布局文件。并在您的公共虚空登录中添加此

请记住：直接将其放在空白中而不是if语句

中

Log.i("LOGIN DETAILS","Username: "+username.getText().toString()+"Password: "+password.getText().toString());

然后使用信息过滤器发布您的日志猫。

如果对日志猫不太好，您可以在公共无效登录中添加此行代码而不是if语句

Toast.makeText(login.this, "Username: "+ username.getText().toString() + "password: " + password.getText().toString(), Toast.LENGTH_LONG).show();

Android Studio - 登录页面无效

1 个答案: