android sqllite登录

以下是一个简单的Android SQLite登录的例子：

public class LoginActivity extends AppCompatActivity {
    private EditText mUsernameEditText;
    private EditText mPasswordEditText;
    private Button mLoginButton;
    private SQLiteDatabase mDatabase;

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

        // 初始化UI组件
        mUsernameEditText = findViewById(R.id.username_edit_text);
        mPasswordEditText = findViewById(R.id.password_edit_text);
        mLoginButton = findViewById(R.id.login_button);

        // 打开或创建数据库
        mDatabase = openOrCreateDatabase("user.db", MODE_PRIVATE, null);

        // 创建用户表
        mDatabase.execSQL("CREATE TABLE IF NOT EXISTS user (id INTEGER PRIMARY KEY AUTOINCREMENT, name TEXT, password TEXT)");

        // 添加测试用户
        mDatabase.execSQL("INSERT INTO user (name, password) VALUES ('test', '123456')");

        // 设置登录按钮的点击事件
        mLoginButton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                String username = mUsernameEditText.getText().toString();
                String password = mPasswordEditText.getText().toString();

                // 查询用户是否存在
                Cursor cursor = mDatabase.rawQuery("SELECT * FROM user WHERE name = ? AND password = ?", new String[]{username, password});

                if (cursor.moveToFirst()) {
                    // 登录成功
                    Toast.makeText(LoginActivity.this, "登录成功", Toast.LENGTH_SHORT).show();
                } else {
                    // 登录失败
                    Toast.makeText(LoginActivity.this, "用户名或密码错误", Toast.LENGTH_SHORT).show();
                }

                cursor.close();
            }
        });
    }
}