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Swift DateComponentsFormatter drop leading zeroes but keep at least one digit in Minutes place

I am using the following DateComponentsFormatter in Swift:

let formatter = DateComponentsFormatter()
formatter.unitsStyle   = .positional
formatter.allowedUnits = [.hour, .minute, .second]
formatter.zeroFormattingBehavior = [.default]

This produces results like 2:41 (for 2 minutes and 41 seconds) and 08 (for 8 seconds). I would, however, like to keep at least one digit in the Minutes place, returning 0:08 instead of 08. Is this possible with the DateComponentsFormatter? I would prefer a solution that returns a localized result.

like image 576
Evan Kaminsky Avatar asked Feb 12 '19 01:02

Evan Kaminsky


2 Answers

Yes. This can be easily accomplished adding a ternary conditional based on the time interval when setting the date components formatter allowed units:


Xcode 11 • Swift 5.1

extension Formatter {
    static let positional: DateComponentsFormatter = {
        let positional = DateComponentsFormatter()
        positional.unitsStyle = .positional
        positional.zeroFormattingBehavior = .pad
        return positional
    }()
}

extension TimeInterval {
    var positionalTime: String {
        Formatter.positional.allowedUnits = self >= 3600 ?
                                            [.hour, .minute, .second] :
                                            [.minute, .second]
        let string = Formatter.positional.string(from: self)!
        return string.hasPrefix("0") && string.count > 4 ?
            .init(string.dropFirst()) : string
    }
}

Usage

8.0.positionalTime     //    "0:08"
161.0.positionalTime   //    "2:41"
3600.0.positionalTime  // "1:00:00"
like image 52
Leo Dabus Avatar answered Oct 16 '22 03:10

Leo Dabus


Came up with this solution, employing a regex, that seemingly covers all cases:

let formatter = DateComponentsFormatter()
formatter.allowedUnits = [.hour, .minute, .second]
formatter.unitsStyle = .positional
formatter.zeroFormattingBehavior = .pad

func durationString(for duration: TimeInterval) -> String {
    formatter.string(from: duration)!
        .replacingOccurrences(of: #"^00[:.]0?|^0"#, with: "", options: .regularExpression)
}

durationString(for: 0)      // 0:00
durationString(for: 1)      // 0:01
durationString(for: 10)     // 0:10
durationString(for: 60)     // 1:00
durationString(for: 600)    // 10:00
durationString(for: 3600)   // 1:00:00
durationString(for: 36000)  // 10:00:00
like image 2
Xtian D. Avatar answered Oct 16 '22 02:10

Xtian D.