210

I have a (somewhat?) basic question regarding time conversions in Swift.

I have an integer that I would like converted into Hours / Minutes / Seconds.

Example: Int = 27005 would give me:

7 Hours  30 Minutes 5 Seconds

I know how to do this in PHP, but alas, swift isn't PHP.

SecretAgentMan
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Joe
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25 Answers25

401

Define

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

Use

> secondsToHoursMinutesSeconds(27005)
(7,30,5)

or

let (h,m,s) = secondsToHoursMinutesSeconds(27005)

The above function makes use of Swift tuples to return three values at once. You destructure the tuple using the let (var, ...) syntax or can access individual tuple members, if need be.

If you actually need to print it out with the words Hours etc then use something like this:

func printSecondsToHoursMinutesSeconds(_ seconds: Int) {
  let (h, m, s) = secondsToHoursMinutesSeconds(seconds)
  print ("\(h) Hours, \(m) Minutes, \(s) Seconds")
}

Note that the above implementation of secondsToHoursMinutesSeconds() works for Int arguments. If you want a Double version you'll need to decide what the return values are - could be (Int, Int, Double) or could be (Double, Double, Double). You could try something like:

func secondsToHoursMinutesSeconds(seconds: Double) -> (Double, Double, Double) {
  let (hr,  minf) = modf(seconds / 3600)
  let (min, secf) = modf(60 * minf)
  return (hr, min, 60 * secf)
}
aheze
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GoZoner
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    The last value `(seconds % 3600) % 60` can be optimized to `seconds % 60`. No need to extract the hours first. – zisoft Nov 07 '14 at 07:02
  • @GoZoner - i can't seem to be getting the printSecondsToHoursMinutesSeconds function to be working properly. This is what I have in a playground, but printSecondsToHoursMinutesSeconds isnt returning anything: import UIKit func secondsToHoursMinutesSeconds (seconds : Int) -> (Int, Int, Int) { return (seconds / 3600, (seconds % 3600) / 60, (seconds % 3600) % 60) } let (h,m,s) = secondsToHoursMinutesSeconds(27005) func printSecondsToHoursMinutesSeconds (seconds:Int) -> () { let (h, m, s) = secondsToHoursMinutesSeconds (seconds) println ("\(h) Hours, \(m) Minutes, \(s) Seconds") } – Joe Nov 10 '14 at 02:26
  • `printSecondstoHoursMinutesSeconds()` doesn't return anything (see the `-> ()` return type in the function declaration). The function prints something; which does not show up in a Playground. If you want it to return something, say a `String` then eliminate the `println()` call and fix the function's return type. – GoZoner Nov 10 '14 at 20:58
  • @GoZoner - just a quick question in your syntax above. How would I declare that 27005 in a variable? I'm working on a tool right now to get my feet wet with swift. I have a constant that displays the amount of seconds (generated after my basic calculations). let cocSeconds = cocMinutes * 60. (cocSeconds is what I want to use in place of 27005.) I think the problem I am having is cocSeconds is a double, and in your syntax, you're using Ints. How would I go about adjusting this code to put a variable in place of 27005? Thank you so much in advance!!! – Joe Nov 29 '14 at 03:56
  • The solution I gave works for `Int` types because of the nature of the `/` and `%` functions. That is `/` drops the factional part. The same code won't work for `Double`. Specifically 2 == 13/5 but 2.6 == 13.0/5. Therefore you would need a different implementation for `Double`. I've updated my answer with a note on this. – GoZoner Nov 29 '14 at 05:04
271

In macOS 10.10+ / iOS 8.0+ (NS)DateComponentsFormatter has been introduced to create a readable string.

It considers the user's locale und language.

let interval = 27005

let formatter = DateComponentsFormatter()
formatter.allowedUnits = [.hour, .minute, .second]
formatter.unitsStyle = .full

let formattedString = formatter.string(from: TimeInterval(interval))!
print(formattedString)

The available unit styles are positional, abbreviated, short, full, spellOut and brief.

For more information please read the documenation.

vadian
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118

Building upon Vadian's answer, I wrote an extension that takes a Double (of which TimeInterval is a type alias) and spits out a string formatted as time.

extension Double {
  func asString(style: DateComponentsFormatter.UnitsStyle) -> String {
    let formatter = DateComponentsFormatter()
    formatter.allowedUnits = [.hour, .minute, .second, .nanosecond]
    formatter.unitsStyle = style
    return formatter.string(from: self) ?? ""
  }
}

Here are what the various DateComponentsFormatter.UnitsStyle options look like:

10000.asString(style: .positional)  // 2:46:40
10000.asString(style: .abbreviated) // 2h 46m 40s
10000.asString(style: .short)       // 2 hr, 46 min, 40 sec
10000.asString(style: .full)        // 2 hours, 46 minutes, 40 seconds
10000.asString(style: .spellOut)    // two hours, forty-six minutes, forty seconds
10000.asString(style: .brief)       // 2hr 46min 40sec
budiDino
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Adrian
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  • @MaksimKniazev Usually time-oriented values are represented by `Doubles` in Swift. – Adrian Mar 23 '18 at 01:10
  • @Adrian thank you for the extension. I like the .positional UnitStyle, however I would like to display ie: 9 seconds as "00:09" instead of "9", 1min 25 seconds as "01:25" instead of "1:25". I am able to achieve this calculating manually, based on Double value and I was wondering is there a way to incorporate into the extension itself? Thank you. – Vetuka Feb 07 '19 at 22:06
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    FYI: if you want to keep the 0s before a single digit value (or simply if the value is 0) you need to add this `formatter.zeroFormattingBehavior = .pad`. This would give us: `3660.asString(style: .positional) // 01:01:00` – Moumou May 09 '20 at 17:03
  • Note that DateComponentsFormatter isn't available on Linux – Xander Dunn Dec 19 '20 at 13:37
  • @Adrian ou can replace the last two lines with `return formatter.string(from: self) ?? ""` – budiDino Jun 30 '21 at 14:53
  • nanosecond break in swift5 – slboat Dec 17 '21 at 06:49
37

In Swift 5:

    var i = 9897

    func timeString(time: TimeInterval) -> String {
        let hour = Int(time) / 3600
        let minute = Int(time) / 60 % 60
        let second = Int(time) % 60

        // return formated string
        return String(format: "%02i:%02i:%02i", hour, minute, second)
    }

To call function

    timeString(time: TimeInterval(i))

Will return 02:44:57

DialDT
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31

I have built a mashup of existing answers to simplify everything and reduce the amount of code needed for Swift 3.

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)"
}

Usage:

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)")                
}

Prints:

00:01:40

Community
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David Seek
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28

Here is a more structured/flexible approach: (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)

Having an approach like this allows the adding of convenience parsing like this:

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

It should be noted that purely Integer based approaches don't take leap day/seconds into account. If the use case is dealing with real dates/times Date and Calendar should be used.

GetSwifty
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  • Could you please add `month` to your implementation? Thanks in advance! – ixany Dec 05 '16 at 15:31
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    You should use NSCalendar (Calendar) for something like that. – GetSwifty Dec 05 '16 at 16:08
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    You could replace `return number < 10 ? "0\(number)" : "\(number)"` by using a string formatter `return String(format: "%02d", number)` which automatically adds a zero when the number is below 10 – Continuum Jan 10 '20 at 01:28
19

Swift 5:

extension Int {

    func secondsToTime() -> String {

        let (h,m,s) = (self / 3600, (self % 3600) / 60, (self % 3600) % 60)

        let h_string = h < 10 ? "0\(h)" : "\(h)"
        let m_string =  m < 10 ? "0\(m)" : "\(m)"
        let s_string =  s < 10 ? "0\(s)" : "\(s)"

        return "\(h_string):\(m_string):\(s_string)"
    }
}

Usage:

let seconds : Int = 119
print(seconds.secondsToTime()) // Result = "00:01:59"
ZAFAR007
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15

Xcode 12.1. Swift 5

DateComponentsFormatter: A formatter that creates string representations, by using unitsStyle u can get a string as you want and mention allowedUnits.

e.g: output for unitsStyle:: for 10000 secods

  1. full = "2 hours, 46 minutes, 49 seconds"
  2. positional = "2:46:40"
  3. abbreviated = "2h 46m 40s"
  4. spellOut = "two hours, forty-six minutes, forty seconds”
  5. short = "2hr,46 min,40 sec"
  6. brief = "2hr 46min 40sec"

Easy to use:

 let time = convertSecondsToHrMinuteSec(seconds: 10000)


func convertSecondsToHrMinuteSec(seconds:Int) -> String{
     let formatter = DateComponentsFormatter()
     formatter.allowedUnits = [.hour, .minute, .second]
     formatter.unitsStyle = .full
    
     let formattedString = formatter.string(from:TimeInterval(seconds))!
     print(formattedString)
     return formattedString
    }
iOS_Tejas
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14

Swift 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)
}
r3dm4n
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  • What is truncatingRemainder ? Xcode does not recognize it for me. – Ahmadreza Jul 08 '19 at 07:31
  • you can read more about truncatingRemainder [here](https://developer.apple.com/documentation/swift/double/2885641-truncatingremainder) – r3dm4n Mar 23 '21 at 14:49
13

Here is another simple implementation in Swift3.

func seconds2Timestamp(intSeconds:Int)->String {
   let mins:Int = intSeconds/60
   let hours:Int = mins/60
   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
}
neeks
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12

SWIFT 3.0 solution based roughly on the one above using extensions.

extension CMTime {
  var durationText:String {
    let totalSeconds = CMTimeGetSeconds(self)
    let hours:Int = Int(totalSeconds.truncatingRemainder(dividingBy: 86400) / 3600)
    let minutes:Int = Int(totalSeconds.truncatingRemainder(dividingBy: 3600) / 60)
    let seconds:Int = Int(totalSeconds.truncatingRemainder(dividingBy: 60))

    if hours > 0 {
        return String(format: "%i:%02i:%02i", hours, minutes, seconds)
    } else {
        return String(format: "%02i:%02i", minutes, seconds)
    }

  }
}

Use it with AVPlayer calling it like this?

 let dTotalSeconds = self.player.currentTime()
 playingCurrentTime = dTotalSeconds.durationText
user3069232
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8

I had answered to the similar question, however you don't need to display milliseconds in the result. Hence my solution requires iOS 10.0, tvOS 10.0, watchOS 3.0 or macOS 10.12.

You should call func convertDurationUnitValueToOtherUnits(durationValue:durationUnit:smallestUnitDuration:) from the answer that I already mentioned here:

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
Roman Podymov
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6

Answer of @r3dm4n was great. However, I also needed hour in it. Just in case someone else needed too here it is:

func formatSecondsToString(_ seconds: TimeInterval) -> String {
    if seconds.isNaN {
        return "00:00:00"
    }
    let sec = Int(seconds.truncatingRemainder(dividingBy: 60))
    let min = Int(seconds.truncatingRemainder(dividingBy: 3600) / 60)
    let hour = Int(seconds / 3600)
    return String(format: "%02d:%02d:%02d", hour, min, sec)
}
Vetuka
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5

Swift 5 & String Response, In presentable format

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)
 }

Usage:

print(secondsToHoursMinutesSecondsStr(seconds: 20000)) // Result = "5 h 33 min 20 sec"
Asad Ali Choudhry
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3

According to GoZoner answer I have wrote an Extension to get the time formatted according to the hours, minute, and seconds:

extension Double {

    func secondsToHoursMinutesSeconds () -> (Int?, Int?, Int?) {
        let hrs = self / 3600
        let mins = (self.truncatingRemainder(dividingBy: 3600)) / 60
        let seconds = (self.truncatingRemainder(dividingBy:3600)).truncatingRemainder(dividingBy:60)
        return (Int(hrs) > 0 ? Int(hrs) : nil , Int(mins) > 0 ? Int(mins) : nil, Int(seconds) > 0 ? Int(seconds) : nil)
    }

    func printSecondsToHoursMinutesSeconds () -> String {

        let time = self.secondsToHoursMinutesSeconds()

        switch time {
        case (nil, let x? , let y?):
            return "\(x) min \(y) sec"
        case (nil, let x?, nil):
            return "\(x) min"
        case (let x?, nil, nil):
            return "\(x) hr"
        case (nil, nil, let x?):
            return "\(x) sec"
        case (let x?, nil, let z?):
            return "\(x) hr \(z) sec"
        case (let x?, let y?, nil):
            return "\(x) hr \(y) min"
        case (let x?, let y?, let z?):
            return "\(x) hr \(y) min \(z) sec"
        default:
            return "n/a"
        }
    }
}

let tmp = 3213123.printSecondsToHoursMinutesSeconds() // "892 hr 32 min 3 sec"
xGoPox
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3

Here is what I use for my Music Player in Swift 4+. I am converting seconds Int to readable String format

extension Int {
    var toAudioString: String {
        let h = self / 3600
        let m = (self % 3600) / 60
        let s = (self % 3600) % 60
        return h > 0 ? String(format: "%1d:%02d:%02d", h, m, s) : String(format: "%1d:%02d", m, s)
    }
}

Use like this:

print(7903.toAudioString)

Output: 2:11:43

Yaroslav Dukal
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3

Latest Code: XCode 10.4 Swift 5

extension Int {
    func timeDisplay() -> String {
        return "\(self / 3600):\((self % 3600) / 60):\((self % 3600) % 60)"
    }
}
Saranjith
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3

From @Gamec answer

typealias CallDuration = Int

extension CallDuration {
    func formatedString() -> String? {
        let hours = self / 3600
        let minutes = (self / 60) % 60
        let seconds = self % 60
        if hours > 0 { return String(format: "%0.2d:%0.2d:%0.2d", hours, minutes, seconds) }
        return String(format: "%0.2d:%0.2d", minutes, seconds)
    }
}


Usage:

let duration: CallDuration = 3000
duration.formatedString() // 50 minute
2

The simplest way imho:

let hours = time / 3600
let minutes = (time / 60) % 60
let seconds = time % 60
return String(format: "%0.2d:%0.2d:%0.2d", hours, minutes, seconds)
Gamec
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1

NSTimeInterval is Double do do it with extension. Example:

extension Double {

    var formattedTime: String {

        var formattedTime = "0:00"

        if self > 0 {

            let hours = Int(self / 3600)
            let minutes = Int(truncatingRemainder(dividingBy: 3600) / 60)

            formattedTime = String(hours) + ":" + (minutes < 10 ? "0" + String(minutes) : String(minutes))
        }

        return formattedTime
    }
}
Bartłomiej Semańczyk
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1

convert a number into time as a string

func convertToHMS(number: Int) -> String {
  let hour    = number / 3600;
  let minute  = (number % 3600) / 60;
  let second = (number % 3600) % 60 ;
  
  var h = String(hour);
  var m = String(minute);
  var s = String(second);
  
  if h.count == 1{
      h = "0\(hour)";
  }
  if m.count == 1{
      m = "0\(minute)";
  }
  if s.count == 1{
      s = "0\(second)";
  }
  
  return "\(h):\(m):\(s)"
}
print(convertToHMS(number:3900))
Sami
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0

I went ahead and created a closure for this (in Swift 3).

let (m, s) = { (secs: Int) -> (Int, Int) in
        return ((secs % 3600) / 60, (secs % 3600) % 60) }(299)

This will give m = 4 and s = 59. So you can format that as you wish. You may of course want to add hours as well, if not more information.

0

Swift 4 I'm using this extension

 extension Double {

    func stringFromInterval() -> String {

        let timeInterval = Int(self)

        let millisecondsInt = Int((self.truncatingRemainder(dividingBy: 1)) * 1000)
        let secondsInt = timeInterval % 60
        let minutesInt = (timeInterval / 60) % 60
        let hoursInt = (timeInterval / 3600) % 24
        let daysInt = timeInterval / 86400

        let milliseconds = "\(millisecondsInt)ms"
        let seconds = "\(secondsInt)s" + " " + milliseconds
        let minutes = "\(minutesInt)m" + " " + seconds
        let hours = "\(hoursInt)h" + " " + minutes
        let days = "\(daysInt)d" + " " + hours

        if daysInt          > 0 { return days }
        if hoursInt         > 0 { return hours }
        if minutesInt       > 0 { return minutes }
        if secondsInt       > 0 { return seconds }
        if millisecondsInt  > 0 { return milliseconds }
        return ""
    }
}

useage

// assume myTimeInterval = 96460.397    
myTimeInteval.stringFromInterval() // 1d 2h 47m 40s 397ms
caesss
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0

neek's answer isn't correct.

here's the correct version

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
}
0

Another way would be convert seconds to date and take the components i.e seconds, minutes and hour from date itself. This solution has limitation only till 23:59:59

Wasim
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