Swift 5 &字符串响应，在像样的格式

public static func secondsToHoursMinutesSecondsStr (seconds : Int) -> String {
      let (hours, minutes, seconds) = secondsToHoursMinutesSeconds(seconds: seconds);
      var str = hours > 0 ? "\(hours) h" : ""
      str = minutes > 0 ? str + " \(minutes) min" : str
      str = seconds > 0 ? str + " \(seconds) sec" : str
      return str
  }

public static func secondsToHoursMinutesSeconds (seconds : Int) -> (Int, Int, Int) {
        return (seconds / 3600, (seconds % 3600) / 60, (seconds % 3600) % 60)
 }

用法:

print(secondsToHoursMinutesSecondsStr(seconds: 20000)) // Result = "5 h 33 min 20 sec"

我已经构建了一个现有答案的mashup，以简化一切并减少Swift 3所需的代码量。

func hmsFrom(seconds: Int, completion: @escaping (_ hours: Int, _ minutes: Int, _ seconds: Int)->()) {

        completion(seconds / 3600, (seconds % 3600) / 60, (seconds % 3600) % 60)

}

func getStringFrom(seconds: Int) -> String {

    return seconds < 10 ? "0\(seconds)" : "\(seconds)"
}

用法:

var seconds: Int = 100

hmsFrom(seconds: seconds) { hours, minutes, seconds in

    let hours = getStringFrom(seconds: hours)
    let minutes = getStringFrom(seconds: minutes)
    let seconds = getStringFrom(seconds: seconds)

    print("\(hours):\(minutes):\(seconds)")                
}

打印:

00:01:40

Neek的答案不正确。

这是正确的版本

func seconds2Timestamp(intSeconds:Int)->String {
   let mins:Int = (intSeconds/60)%60
   let hours:Int = intSeconds/3600
   let secs:Int = intSeconds%60

   let strTimestamp:String = ((hours<10) ? "0" : "") + String(hours) + ":" + ((mins<10) ? "0" : "") + String(mins) + ":" + ((secs<10) ? "0" : "") + String(secs)
   return strTimestamp
}

斯威夫特4

func formatSecondsToString(_ seconds: TimeInterval) -> String {
    if seconds.isNaN {
        return "00:00"
    }
    let Min = Int(seconds / 60)
    let Sec = Int(seconds.truncatingRemainder(dividingBy: 60))
    return String(format: "%02d:%02d", Min, Sec)
}

我已经回答了类似的问题，但是你不需要在结果中显示毫秒。因此，我的解决方案需要iOS 10.0, tvOS 10.0, watchOS 3.0或macOS 10.12。

你应该调用func convertDurationUnitValueToOtherUnits(durationValue:durationUnit: smallstunitduration:)从我已经在这里提到的答案:

let secondsToConvert = 27005
let result: [Int] = convertDurationUnitValueToOtherUnits(
    durationValue: Double(secondsToConvert),
    durationUnit: .seconds,
    smallestUnitDuration: .seconds
)
print("\(result[0]) hours, \(result[1]) minutes, \(result[2]) seconds") // 7 hours, 30 minutes, 5 seconds

以下是一个更结构化/灵活的方法:(Swift 3)

struct StopWatch {

    var totalSeconds: Int

    var years: Int {
        return totalSeconds / 31536000
    }

    var days: Int {
        return (totalSeconds % 31536000) / 86400
    }

    var hours: Int {
        return (totalSeconds % 86400) / 3600
    }

    var minutes: Int {
        return (totalSeconds % 3600) / 60
    }

    var seconds: Int {
        return totalSeconds % 60
    }

    //simplified to what OP wanted
    var hoursMinutesAndSeconds: (hours: Int, minutes: Int, seconds: Int) {
        return (hours, minutes, seconds)
    }
}

let watch = StopWatch(totalSeconds: 27005 + 31536000 + 86400)
print(watch.years) // Prints 1
print(watch.days) // Prints 1
print(watch.hours) // Prints 7
print(watch.minutes) // Prints 30
print(watch.seconds) // Prints 5
print(watch.hoursMinutesAndSeconds) // Prints (7, 30, 5)

使用这样的方法可以添加方便的解析，如下所示:

extension StopWatch {

    var simpleTimeString: String {
        let hoursText = timeText(from: hours)
        let minutesText = timeText(from: minutes)
        let secondsText = timeText(from: seconds)
        return "\(hoursText):\(minutesText):\(secondsText)"
    }

    private func timeText(from number: Int) -> String {
        return number < 10 ? "0\(number)" : "\(number)"
    }
}
print(watch.simpleTimeString) // Prints 07:30:05

值得注意的是，纯基于整数的方法不考虑闰日/秒。如果用例处理的是真实的日期/时间，则应该使用Date和Calendar。