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

0vvn1miw 于 2023-01-18 发布在其他

关注(0)|答案(2)|浏览(114)

我需要在flutter应用程序中创建一个动态命名数据库的连接，我如何将参数传递给一个具有数据库名称的类，并返回一个连接？
我知道getter不能接受参数，我尝试添加setter函数来设置_databaseName，但没有成功。
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,
    );
  }
}

我成功地调用了它，使用的函数如下：

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

我想打这样的电话

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

dart

来源：https://stackoverflow.com/questions/75156331/flutter-dart-sqflite-how-to-pass-a-dynamic-database-name-to-a-class-to-return-d

2条答案

按热度按时间

llmtgqce1#

这是否可行：

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,
    );
  }
}

赞(0）回复(0）举报 2023-01-18

q8l4jmvw2#

然后使用method而不是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,
  );
}

赞(0）回复(0）举报 2023-01-18

我来回答

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

2条答案

相关问题

热门标签

最新问答