Question

我一直试图制作一个登录应用。这是LoginActivity：

public class LoginActivity extends AppCompatActivity {

EditText UsernameLogin, PasswordLogin;
UserDbHelper userDbHelper;
SQLiteDatabase sqLiteDatabase;
public static Cursor cursor;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.layout_login_activity);
    Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
    setSupportActionBar(toolbar);

}

public void loginUser(View view) {

    UsernameLogin = (EditText) findViewById(R.id.insert_username);
    PasswordLogin = (EditText) findViewById(R.id.insert_password);
    String Username = UsernameLogin.getText().toString();
    String Password = PasswordLogin.getText().toString();

    userDbHelper = new UserDbHelper(getApplicationContext());
    sqLiteDatabase = userDbHelper.getReadableDatabase();
    cursor = userDbHelper.getUsername(Username, sqLiteDatabase);

    if(cursor.getCount() != 0) {
        if(cursor.getString(3).equals(Password)) {
            Intent intent = new Intent(this, WelcomeActivity.class);
            startActivity(intent);
            Toast.makeText(this, "Login successful!", Toast.LENGTH_LONG).show();
        }
        else {
            Toast.makeText(this, "Wrong Password", Toast.LENGTH_SHORT).show();
            PasswordLogin.setText("");
        }
    } else {
        Toast.makeText(this, "Username does not exist", Toast.LENGTH_SHORT).show();
        UsernameLogin.setText("");
        PasswordLogin.setText("");
    }

}

}

这是getUsername内的UserDbHelper方法：

public Cursor getUsername (String username, SQLiteDatabase db) {

    String[] projections = {UserContract.NewUserInfo.FIRST_NAME, UserContract.NewUserInfo.LAST_NAME,
            UserContract.NewUserInfo.USER_PASSWORD, UserContract.NewUserInfo.USER_EMAIL};
    String selection = UserContract.NewUserInfo.USER_NAME + " LIKE ?";
    String[] selection_args = {username};

    Cursor cursor = db.query(UserContract.NewUserInfo.TABLE_NAME, projections, selection, selection_args, null, null, null);

    return cursor;
}

这会导致我的应用崩溃。如果我用字符串替换cursor.getString(3)，则应用程序可以正常运行。

Answer 1

不要使用魔法数字。 Cursor有方法getColumnIndex例如

cursor.getString(cursor.getColumnIndex(UserContract.NewUserInfo.USER_PASSWORD));

并且忘记在查询之前忘记致电cursor.moveToFirst()

使用cursor.getString（）会崩溃我的应用程序

1 个答案: