Question

登录后，我想显示一条消息“欢迎，（用户名）！”在下一个活动。现在我完全不知道如何实现它。那么有人可以给我一个建议吗？

到目前为止，我只是设法实施了注册＆amp;登录部分。

以下是存储用户数据并启用登录的代码段。

MembershipDatabaseHelper.java

public final class MembershipDatabaseHelper extends SQLiteOpenHelper {
    private static final String DB_NAME = "members.db";
    private static final String DB_TABLE = "members";
    private static final int DB_VERSION = 1;
    private static final String COLUMN_ID = "_id";
    private static final String COLUMN_EMAIL = "email";
    private static final String COLUMN_FIRSTNAME = "firstname";
    private static final String COLUMN_LASTNAME = "lastname";
    private static final String COLUMN_PASSWORD = "password";
    private static final String COLUMN_DOB = "dob";
    private static final String CREATE_TABLE = "CREATE TABLE " + DB_TABLE + " (_id INTEGER PRIMARY KEY AUTOINCREMENT, "
    + COLUMN_EMAIL + " TEXT NOT NULL, " + COLUMN_FIRSTNAME + " TEXT NOT NULL, " + COLUMN_LASTNAME + " TEXT NOT NULL, "
    + COLUMN_PASSWORD + " TEXT NOT NULL, " + COLUMN_DOB + " TEXT NOT NULL);";

    private static String queryDrop = "DROP TABLE IF EXISTS " + DB_TABLE;

    SQLiteDatabase db;

    public MembershipDatabaseHelper(Context context) {
        super(context, DB_NAME, null, DB_VERSION);
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        db.execSQL(CREATE_TABLE);
        this.db = db;
    }

    public void signup(User user) {
        db = this.getWritableDatabase();
        ContentValues values = new ContentValues();

        String query = "SELECT * FROM " + DB_TABLE;
        Cursor cursor = db.rawQuery(query, null);
        int count = cursor.getCount();

        values.put(COLUMN_ID, count);
        values.put(COLUMN_EMAIL, user.getEmail());
        values.put(COLUMN_FIRSTNAME, user.getFirstname());
        values.put(COLUMN_LASTNAME, user.getLastname());
        values.put(COLUMN_PASSWORD, user.getPassword());
        values.put(COLUMN_DOB, user.getDob());
        db.insert(DB_TABLE, null, values);
        db.close();
    }

    public String searchPassword(String email) {
        db = this.getReadableDatabase();
        String query = "SELECT " + COLUMN_EMAIL + ", " + COLUMN_PASSWORD + " FROM " + DB_TABLE;
        Cursor cursor = db.rawQuery(query, null);

        String targetEmail;
        String password = "NOT FOUND";

        if(cursor.moveToFirst()) {
            do {
                targetEmail = cursor.getString(0);

                if(targetEmail.equals(email)) {
                    password = cursor.getString(1);
                    break;
                }
            } while(cursor.moveToNext());
        }
        return password;
    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        db.execSQL(queryDrop);
        this.onCreate(db);
    }
}

这是登录部分。

public class LoginActivity extends Activity implements OnClickListener {

    Button login;
    EditText email, password;
    TextView signuphere;
    MembershipDatabaseHelper dbHelper = new MembershipDatabaseHelper(this);

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

        email = (EditText) findViewById(R.id.enter_email);
        password = (EditText) findViewById(R.id.enter_password);
        login = (Button) findViewById(R.id.login);
        signuphere = (TextView) findViewById(R.id.signuphere);

        login.setOnClickListener(this);
        signuphere.setOnClickListener(this);
    }

    @Override
    public void onClick(View v) {
        switch(v.getId()) {
        case(R.id.login):

            String enteredEmail = email.getText().toString();
            String enteredPassword = password.getText().toString();
            String storedPassword = dbHelper.searchPassword(enteredEmail);

            if(enteredPassword.equals(storedPassword)) {
                Toast.makeText(this, "Successfully logged in!", Toast.LENGTH_SHORT).show();
                Intent intent = new Intent(LoginActivity.this, MainScreenActivity.class);
                startActivity(intent);
                finish();
            } else {
                Toast.makeText(this, "Email and password don't match.", Toast.LENGTH_SHORT).show();
            }
            break;

        case(R.id.signuphere):
            startActivity(new Intent(this, SignupActivity.class));
            break;
        }
    }

    @Override
    protected void onDestroy() {
        super.onDestroy();
    }

    @Override
    public void onBackPressed() {
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        builder.setTitle("You haven't finished yet!").setMessage("Are you sure to quit?").setCancelable(false).setPositiveButton("Yes", new DialogInterface.OnClickListener() {

            @Override
            public void onClick(DialogInterface dialog, int which) {
                this.close();
            }

            private void close() {
                finish();
            }   
        }).setNegativeButton("No", new DialogInterface.OnClickListener() {

            @Override
            public void onClick(DialogInterface dialog, int which) {
                dialog.cancel();    
            }
        });

        AlertDialog dialog = builder.create();
        dialog.show();
    }
}

Answer 1

您可以将共享首选项用于此目的。 SharedPreferences是Android SDK中用于存储和检索应用程序首选项的API。 SharedPreferences只是持久存储的数据值集合，您还可以在应用程序包中的所有活动中存储数据。

当应用使用单点登录时，使用SharedPreferences存储access_token，access_expire，username等。完成使用应用程序登录后，它会将一些数据（包括access_token和access_expire）存储到应用程序首选项中。当你重新打开时，它将检查access_token和access_expire值，如果access_token返回null，则会话不再有效，你必须再次登录。

        <?php
      if($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST)) 
       {

          $login    = $_POST['login'];
          $password = $_POST['password'];
          $email    = $_POST['email'];

        if(empty($login) || empty($password) || empty($email)) // Sprawdzanie czy pola formularza nie są puste
        {
          die('Wypełnij wszystkie dane.');
        }
        elseif(!filter_var($email, FILTER_VALIDATE_EMAIL)) // Sprawdzanie poprawności adresu Email
        {
         die('Nie poprawny adres E-mail.');
        }
        else
        {   
         include_once('config.php');

         $mysqli = new mysqli('mysql.hostinger.pl', 'u162600159_reg', '******') or die ('Fatal Error: '.mysql_error());

         if($mysqli -> connect_error) 
          die('Problem z połączeniem się z bazą danych:'.$mysqli -> connect_error.'['.$mysqli -> connect_errno.']');


         $login     = trim(htmlspecialchars($mysqli -> real_escape_string($login)));
         $password  = hash('sha256', trim(htmlspecialchars($mysqli -> real_escape_string($password))));
         $email     = trim(htmlspecialchars($mysqli -> real_escape_string($email)));
         $ip        = $_SERVER['REMOTE_ADDR'];


         $stmt = $mysqli -> prepare("INSERT INTO `user`(`id_user`, `login`,`password`,`email`,`added`,`ip`) VALUES('', ? , ? , ? , now(), ?)");
         $stmt->bind_param("ssss", $login, $password, $email, $ip);

          echo 'Zostałeś pomyślnie zarejestrowany';
         }
        }


    ?>
 </section> 
</body>
</html>

希望这会对你有所帮助。

Answer 2

只需将其与意图一起发送，假设用户名存储在名为username的变量中，并且您的意图称为welcomeIntent，您可以使用putExtra()：

welcomeIntent.putExtra("username",username);

并在您的活动中使用：

Intent.getStringExtra("username");

顺便检查文档..您可以使用任何密钥而不是"username"

更新

当您在LoginActivity：

中使用代码更新答案时

    if(enteredPassword.equals(storedPassword)) {
      Toast.makeText(this, "Successfully logged in!", Toast.LENGTH_SHORT).show();
      Intent intent = new Intent(LoginActivity.this, MainScreenActivity.class);

            String username = enteredEmail ; // if it's not the mail then grab the username from ur db

            intent.putExtra("USERNAME_KEY",username);
      startActivity(intent);
      finish();
  }

然后在MainScreenActivity使用：

String username = Intent.getStringExtra("USERNAME_KEY");

如何在显示屏上显示当前用户的姓名？

2 个答案: