Flutter/Dart/sqflite:如何将动态数据库名称传递给类以返回数据库连接

Question

我需要创建一个连接到一个动态命名的数据库在一个扑动应用程序，我将如何传递一个参数与数据库的名称类，并返回一个连接到它?

我知道getter不能接受参数，我尝试添加setter函数来设置_databaseName是不成功的。

我有一个数据库类如下:

import 'package:path/path.dart';
import 'package:sqflite/sqflite.dart';
import 'package:path/path.dart' show join;
import 'dart:io' as io;
import 'package:path_provider/path_provider.dart';
class DatabaseHelper {
static const _databaseName = "tinkershell.db";
static const _databaseVersion = 1;
DatabaseHelper._privateConstructor(); // make this a singleton class
static final DatabaseHelper instance = DatabaseHelper._privateConstructor();
static Database?  _database; // only have a single app-wide reference to the database
static String path = "";
Future<Database> get database async => _database ??= await _initDatabase();
_onOpen(Database db) async {
//do something here later
}
Future<Database> _initDatabase() async {
io.Directory documentDirectory = await getApplicationDocumentsDirectory();
path = join(documentDirectory.path , _databaseName);
return await openDatabase(
join(path),
version: _databaseVersion,
onOpen: _onOpen,
);
}
}

，我正在使用下面的函数成功地调用它:

Future<void> createNewApplication() async {
Database db = await DatabaseHelper.instance.database;
final data = db.rawQuery(" SELECT * from myTable");
}

我想打这样一个电话:

Future<void> createNewApplication() async {
Database db = await DatabaseHelper.instance.database("database1234");
final data = db.rawQuery(" SELECT * from myTable");
}

Answer 1

这样行吗?

import 'package:path/path.dart';
import 'package:sqflite/sqflite.dart';
import 'package:path/path.dart' show join;
import 'dart:io' as io;
import 'package:path_provider/path_provider.dart';
class DatabaseHelper {
static const _databaseName = "tinkershell.db";
static const _databaseVersion = 1;
DatabaseHelper._privateConstructor(); // make this a singleton class
static final DatabaseHelper instance = DatabaseHelper._privateConstructor();
static Database?  _database; // only have a single app-wide reference to the database
static String path = "";
Future<Database> database(String databaseName) async {
if (_databaseName==databaseName) return _database ??= await _initDatabase(databaseName);

_databaseName=databaseName;
return await _initDatabase(databaseName);
}
_onOpen(Database db) async {
//do something here later
}
Future<Database> _initDatabase(String databaseName) async {
io.Directory documentDirectory = await getApplicationDocumentsDirectory();
path = join(documentDirectory.path , databaseName);
return await openDatabase(
join(path),
version: _databaseVersion,
onOpen: _onOpen,
);
}
}

Answer 2

然后使用method代替getter并传递数据库名称，如下所示:

Future<Database> database(String databaseName) async =>_database ??= await _initDatabase(databaseName);
Future<Database> _initDatabase(String databaseName) async {
io.Directory documentDirectory = await getApplicationDocumentsDirectory();
path = join(documentDirectory.path, databaseName); // use passed databaseName
return await openDatabase(
join(path),
version: _databaseVersion,
onOpen: _onOpen,
);
}