如何在 flutter 中明智地获取 SQFlite 数据日期？

Question

我正在用 flutter 创建待办事项应用程序。我需要明智地显示 Todos 日期。就像今天创建的所有 Todos 都应该显示在 Today 下一样，所有明天的 Todos 都应该显示在 Tomorrow 下。

我是这样创建我的 table 的：

database.execute("""
          CREATE TABLE Todotable(
            id INTEGER PRIMARY KEY AUTOINCREMENT,
            taskName TEXT NOT NULL,
            taskTag TEXT NOT NULL,
            date TEXT NOT NULL,
            isReminder INTEGER NOT NULL,
            isCompleted INTEGER NOT NULL
          )
     """);

我不知道如何明智地查询 SQFlite 数据并将其格式化为今天和明天。并在今天和明天的部分显示，如设计所示。

感谢您的回答:)

Answer 1

获取明天的待办事项

//for tomorrow
           String tomorrowDate=  DateTime.now().add(Duration(days: 1)).toIso8601String();
              var todosForTomrrow= await database
                        .rawQuery('SELECT * FROM Todotable WHERE date = ?', [tomorrowDate]);
    //for today
               String todayDate=  DateTime.now().toIso8601String();
              var todosForToday= await database
                        .rawQuery('SELECT * FROM Todotable WHERE date = ?', [todayDate]);

日期已转换并以字符串格式保存在此处，并且在插入 table 之前应将日期转换为相同的格式

Answer 2

您可以在 sqflite 中创建 DATETIME 列。这是在 sqflite 数据库 中创建的 Weather table 的示例：

batch.execute('''
CREATE TABLE $tableWeather (
$weatherLocalization TEXT NOT NULL,
$weatherDate DATETIME NOT NULL,
$weatherHumidityPercentage REAL,
$weatherWindDegree REAL,
$weatherWindSpeed REAL,
$weatherPressureHPascal INTEGER,
$weatherTemperatureCelsius REAL,
$weatherDescription TEXT,
$weatherIconId TEXT,
PRIMARY KEY($weatherLocalization, $weatherDate));
''',);

您可以创建一个 Weather 对象，如下所示：

class Weather{
      String localization;
      DateTime date;
      double humidityPercentage;
      double windDegree;
      double windSpeedMS;
      int pressureHPascal;
      double temperatureCelsius;
      String description;
      String iconId;

  Weather({
    @required this.localization,
    @required this.date,
    this.humidityPercentage,
    this.windDegree,
    this.windSpeedMS,
    this.pressureHPascal,
    this.temperatureCelsius,
    this.description,
    this.iconId
  });

  //to be used when inserting a row in the table
  Map<String, dynamic> toMap() {
    final map = new Map<String, dynamic>();
    map["$weatherLocalization"] = localization;
    map["$weatherDate"] = date.toString();
    map["$weatherHumidityPercentage"] = humidityPercentage;
    map["$weatherWindDegree"] = windDegree;
    map["$weatherWindSpeed"] = windSpeedMS;
    map["$weatherPressureHPascal"] = pressureHPascal;
    map["$weatherTemperatureCelsius"] = temperatureCelsius;
    map["$weatherDescription"] = description;
    map["$weatherIconId"] = iconId;
    return map;
  }

  //to be used when converting the row into object
  factory WeatherOnDate.fromMap(Map<String, dynamic> data) => new WeatherOnDate(
      localization: data["$weatherLocalization"],
      date: DateTime.parse(data["$weatherDate"]),
      humidityPercentage: data["$weatherHumidityPercentage"],
      windDegree: data["$weatherWindDegree"],
      windSpeedMS: data["$weatherWindSpeed"],
      pressureHPascal: data["$weatherPressureHPascal"],
      temperatureCelsius: data["$weatherTemperatureCelsius"],
      description: data["$weatherDescription"],
      iconId: data["$weatherIconId"]
  );
}

小心地将您的 DateTime 属性转换为字符串或 int，就像我在 toMap() 函数中所做的那样。

然后，当你想获取一个日期时，你可以这样做：

Future<Weather> fetchWeatherOnDate(DateTime dateTime) async {
  DatabaseHelper _databaseHelper = Injection.injector.get();
  List<Map<String, dynamic>> weatherMaps = await _databaseHelper.db.rawQuery(
      'SELECT * FROM $tableWeather WHERE DATE($weatherDate) = DATE(?)',
      [dateTime.toString()]);

  List<Weather> weathers= [];
  for (final weatherMap in weatherMaps) {
    weathers.add(Weather.fromMap(weatherMap));
  }
  if (weathers.isNotEmpty){
    return weathers[0];
  }
  return null;
}

DateTime today = DateTime.now()
Weather weatherToday = fetchWeatherOnDate(today);

我认为它可以让您很好地了解如何解决您的问题:)

Answer 3

Flutter SQLite 包不支持 DateTime。检查有关支持的 SQLite 类型的信息： https://pub.dev/packages/sqflite#supported-sqlite-types

DateTime is not a supported SQLite type. Personally I store them as int (millisSinceEpoch) or string (iso8601)

我们可以按照包开发人员的建议使用 String 或 int。要在数据库中有条件地选择当天的项目，也许字符串更方便。

个人比较喜欢用user formatter来获取搜索关键词，如下图：

    DateFormat formater = DateFormat('yyyyMMdd');
    var today = formater.format(DateTime.now());
    var tomorrow = formater.format(DateTime.now().add(const Duration(days: 1)));

要在数据库中搜索，我会使用如下方法：

Future<TodoList?> getTodoListByDay(String day) async {
    final db = await database;
    String sql =
        "SELECT * FROM Todotable WHERE date = \'${day}\' ";
    var res = await db.rawQuery(sql);
    List<TodoList> objs = res.isNotEmpty
        ? res.map((c) => TodoList.fromMap(c)).toList()
        : [];

    return objs.isNotEmpty ? objs : null;
}

TodoList class 可以从 json 生成，就像我使用 'jsonToDartModel' 包一样。

import 'dart:convert';

class TodoList {
  int? id;
  String? taskName;
  String? taskTag;
  String? date;
  int? isReminder;
  int? isCompleted;

  TodoList({
    this.id,
    this.taskName,
    this.taskTag,
    this.date,
    this.isReminder,
    this.isCompleted,
  });

  factory TodoList.fromMap(Map<String, dynamic> data) => TodoList(
        id: data['id'] as int?,
        taskName: data['taskName'] as String?,
        taskTag: data['taskTag'] as String?,
        date: data['date'] as String?,
        isReminder: data['isReminder'] as int?,
        isCompleted: data['isCompleted'] as int?,
      );

  Map<String, dynamic> toMap() => {
        'id': id,
        'taskName': taskName,
        'taskTag': taskTag,
        'date': date,
        'isReminder': isReminder,
        'isCompleted': isCompleted,
      };

  /// `dart:convert`
  ///
  /// Parses the string and returns the resulting Json object as [TodoList].
  factory TodoList.fromJson(String data) {
    return TodoList.fromMap(json.decode(data) as Map<String, dynamic>);
  }

  /// `dart:convert`
  ///
  /// Converts [TodoList] to a JSON string.
  String toJson() => json.encode(toMap());
}

如何在 flutter 中明智地获取 SQFlite 数据日期？

How to get SQFlite data date wise in flutter?

sqlite

dart

flutter

sqflite