我有一个时间间隔,比如12600,相当于3小时30分钟。我如何格式化任何这样的时间间隔,以便仅保留间隔的最高部分(例如,在这种情况下图中,小时)并且将正确的区域设置缩写附加到该数字。例如10米(10分钟),3d(3天),1年(1年)。
编辑:以下是一些例子:
Time interval in: 90000 Whole string: 1d String out: 1d Time interval in: 900 Whole string: 15m String out: 15m Time interval in: 13500 Whole String: 3h 45m String out: 4h
作为一般规则,应用正常的舍入规则(3.4轮向下,3.6轮向上)。
答案 0 :(得分:39)
使用自iOS 8以来可用的DateComponentFormatter
:
func format(duration: TimeInterval) -> String {
let formatter = DateComponentsFormatter()
formatter.allowedUnits = [.day, .hour, .minute, .second]
formatter.unitsStyle = .abbreviated
formatter.maximumUnitCount = 1
return formatter.string(from: duration)!
}
for d in [12600.0, 90000.0, 900.0, 13500.0] {
let str = format(duration: d)
print("\(d): \(str)")
}
打印:
12600.0: 4h
90000.0: 1d
900.0: 15m
13500.0: 4h
答案 1 :(得分:11)
Swift 3 扩展程序:
extension TimeInterval {
func format() -> String? {
let formatter = DateComponentsFormatter()
formatter.allowedUnits = [.day, .hour, .minute, .second, .nanosecond]
formatter.unitsStyle = .abbreviated
formatter.maximumUnitCount = 1
return formatter.string(from: self)
}
}
答案 2 :(得分:6)
万一有人想要它.. Swift 4
extension TimeInterval {
func format(using units: NSCalendar.Unit) -> String? {
let formatter = DateComponentsFormatter()
formatter.allowedUnits = units
formatter.unitsStyle = .abbreviated
formatter.zeroFormattingBehavior = .pad
return formatter.string(from: self)
}}
使用示例:
let value:TimeInterval = 12600.0
print("\(value.format(using: [.hour, .minute, .second])!)")
,结果将是:
3h 30m 0s
答案 3 :(得分:1)
您可以使用NSDate和NSCalendar。你可以这样说:
let timeInterval:Double = 12600
let calendar = NSCalendar.currentCalendar()
let date = NSDate(timeInterval: -timeInterval, sinceDate: NSDate())
let components = calendar.components([.Year,.Day,.Hour, .Minute, .Second, .Nanosecond], fromDate: date, toDate: NSDate(), options: [])
let hour = components.hour //3
let minute = components.minute //30
Per duncan&rmaddy的建议使用NSDateComponentsFormatter
答案 4 :(得分:1)
看一下NSDateComponentsFormatter
课程。它允许您使用2个日期或使用NSTimeInterval计算您想要的任何单位,并自动支持不同的语言和语言环境。在这里有关于这个主题的SO有几个帖子。
答案 5 :(得分:0)
我为你创造了一个功能!我希望你喜欢它。这非常容易实现,并且非常易于定制。
func totime(time: Double) -> (String) {
var timex = time
var fancytime: String = "a while"
if time < 61 {
fancytime = "\(timex)s"
} else if time < 3601 {
timex = timex/60
timex = round(timex)
fancytime = "\(timex)m"
} else if time < 86401 {
timex = timex/3600
timex = round(timex)
fancytime = "\(timex)h"
} else if Double(time) < 3.15576E+07 {
timex = timex/86400
timex = round(timex)
fancytime = "\(timex)d"
} else {
fancytime = "more than one year"
}
fancytime = fancytime.stringByReplacingOccurrencesOfString(".0", withString: "")
return fancytime
}
经过测试,它完美无缺:
print(totime(90000)) // prints "1d"
print(totime(900)) // prints "15m"
print(totime(13500)) // prints "4h"
只需使用totime(Double)
调用它。