Question

我正在 Flutter 中创建 todo 应用程序。我需要明智地显示 Todos 日期。就像今天创建的所有待办事项都应该显示在今天下一样，所有明天的待办事项都应该显示在明天下。

我像这样创建了我的表：

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 数据日期并将其格式化为 Today 和 Tomorrow。并按照设计中所示的今天和明天的部分显示。

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]);

日期被转换并以字符串格式保存在这里，日期在插入表格之前应该转换为相同的格式

Answer 2

您可以在 sqflite 中创建 DATETIME 列。以下是在 sqflite 数据库中创建的 Weather 表的示例：

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));
''',);

您可以按如下方式创建 Weather 对象：

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，当您想要获取日期时，您可以这样做：

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);

我认为它让您对如何解决问题有了很好的了解:)

如何在颤振中明智地获取 SQFlite 数据日期？

2 个答案: