速递支票日期为今天或明天

Question

我有一个整数数组，它是今天和明天日期的集合，我想根据日期的类型来分隔整数数组

let dateCollection = [
1633722900,
1633730500,
1633754910,
1633758913,
1633820400,
1633824000,
1633827600,
1633831200,
1633834800,
1633838400,
1633842000
]

预期结果

let today: [Int] = [
1633722900,
1633730500,
1633754910,
1633758913
]
let tomorrow: [Int] = [  
1633820400,
1633824000,
1633827600,
1633831200,
1633834800,
1633838400,
1633842000
]

我应该做些什么来分开它们，我已经做了一个扩展来将整数转换为日期或反之亦然，并将其显示为时间，我已经创建了日期到时间扩展也

func getTimetringFromINT() -> String {
let date = Date(timeIntervalSince1970: TimeInterval(self))
let dateFormatter = DateFormatter()
dateFormatter.locale = Locale(identifier: "id")
dateFormatter.dateFormat = "HH:mm"
return dateFormatter.string(from: date)
}

选择器，在将日期转换为时间字符串

之后

Answer 1

您可以使用Calendar.current.ordinality来比较不同日期的年份。下面的示例生成今天和明天的日期，然后将它们过滤回单独的数组:

let today = Date()
let todayInts = Array(0..<10).map { today.timeIntervalSince1970 + TimeInterval($0) }
print("Today:", todayInts,"n")
let tomorrow = Calendar.current.date(byAdding: .day, value: 1, to: today)!
let tomorrowInts = Array(0..<10).map { tomorrow.timeIntervalSince1970 + TimeInterval($0) }
print("Tomorrow:", tomorrowInts,"n")
let allInts = todayInts + tomorrowInts
let todayDayOfYear = Calendar.current.ordinality(of: .day, in: .year, for: today)!
let filteredTodayInts = allInts.filter { Calendar.current.ordinality(of: .day, in: .year, for: Date(timeIntervalSince1970: $0)) == todayDayOfYear }
print("Filtered today:", filteredTodayInts,"n")
let tomorrowDayOfYear = todayDayOfYear + 1
let filteredTomorrowInts = allInts.filter { Calendar.current.ordinality(of: .day, in: .year, for: Date(timeIntervalSince1970: $0)) == tomorrowDayOfYear }
print("Filtered tomorrow:", filteredTomorrowInts,"n")

Answer 2

一种有效的方法是计算今天和明天之间午夜的int值，然后根据该值分割数组

let calendar = Calendar.current
var today = [Int]()
var tomorrow = [Int]()
if let midnight = calendar.date(byAdding: .day, value: 1, to: calendar.startOfDay(for: .now))?.timeIntervalSince1970 {
let limit = Int(midnight)
dateCollection.forEach { $0 < limit ? today.append($0) : tomorrow.append($0) }
}

Answer 3

另一种方法是使用CalendarAPI获取第二天的开始日期(不管日光节约时间如何变化)，然后对数组

进行分区

let startOfTomorrow = Calendar.current.nextDate(after: Date(), 
matching: DateComponents(hour: 0), 
matchingPolicy: .nextTime)!
.timeIntervalSince1970
let splitIndex = dateCollection.partition{ $0 < Int(startOfTomorrow) }
let tomorrowDates = dateCollection[..<splitIndex]
let todayDates = dateCollection[splitIndex...]

为了能够运行此命令，将dateCollection声明为var

Answer 4

通过StartOfDay的任何时候,我们可以将所有股票相同StartOfDay日期。

extension Date {
var morning: Date { Calendar.current.startOfDay(for: self) }
var nextMorning: Date { Calendar.current.date(byAdding: .day, value: 1, to: self)!.morning }
}
extension Int {
var since1970: Date { Date(timeIntervalSince1970: TimeInterval(self)) }
}

在本例中:

let todayMorning    = Date().morning     // Date().advanced(by: 3600*24 * -19).morning
let tomorrowMorning = Date().nextMorning // Date().advanced(by: 3600*24 * -19).nextMorning
let dateCollection  = [ ... ]
let today    = dateCollection.filter{ $0.since1970.morning == todayMorning }
let tomorrow = dateCollection.filter{ $0.since1970.morning == tomorrowMorning }