关于NSTimeZone.secondsFromGMT的困惑

Question

我正在开发一个应用程序，有一个功能，进入黑暗/夜间模式在夜间自动。该应用程序询问用户的位置，并使用this algorithm确定日出/日落时间(世界时)。

唯一不清楚的步骤是将UT时间转换为本地时间，因为算法中没有解释这一点。假设我得到的日出时间是8.5 (UT时间早上8:30)。我如何将其转换为用户的本地时间，以检查是白天还是黑夜？或者等效地，我如何将用户的本地时间转换为UT以便能够比较它们？

到目前为止，我已经尝试使用NSCalendar来获取当前日期的NSDateComponents (NSDate())。其中一个组件是我可以从中获取secondsFromGMT的NSTimeZone?。如下所示：

let dateComponents = NSCalendar.currentCalendar().components([.TimeZone], fromDate: NSDate())
let localOffset = Double(dateComponents.timeZone?.secondsFromGMT ?? 0)/3600

其中localOffset应该是从UT (如果我是对的话是GMT )到本地时间的时间差(以小时为单位)，如果是dateComponents.timeZone == nil，则默认为0(我不知道在什么情况下会发生这种情况)。问题是，我现在得到的localOffset比未来6个月的相同(届时夏令时将与我所在的西班牙现在不同)。这是否意味着我需要将daylightSavingTime和/或daylightSavingTimeOffset属性与secondsFromGMT一起使用？这难道不是secondsFromGMT本身造成的吗？

当我读到算法的结果时，事情变得更加令我困惑。日落时间(当地时间)与谷歌给出的时间完全一致，但日出时间比谷歌所说的时间(我的位置和日期)提前了一个小时。我与你分享算法的整个Swift实现，希望它能帮助别人发现我做错了什么。

import Foundation
import CoreLocation

enum SunriseSunsetZenith: Double {
    case Official       =  90.83
    case Civil          =  96
    case Nautical       = 102
    case Astronomical   = 108
}

func sunriseSunsetHoursForLocation(coordinate: CLLocationCoordinate2D, atDate date: NSDate = NSDate(), zenith: SunriseSunsetZenith = .Civil) -> (sunrise: Double, sunset: Double) {
    // Initial values (will be changed later)
    var sunriseTime = 7.5
    var sunsetTime = 19.5

    // Get the longitude and latitude
    let latitude = coordinate.latitude
    let longitude = coordinate.longitude

    // Get the day, month, year and local offset
    let dateComponents = NSCalendar.currentCalendar().components([.Day, .Month, .Year, .TimeZone], fromDate: date)
    let day = Double(dateComponents.day)
    let month = Double(dateComponents.month)
    let year = Double(dateComponents.year)
    let localOffset = Double(dateComponents.timeZone?.daylightSavingTimeOffset ?? 0)/3600

    // Calculate the day of the year
    let N1 = floor(275*month/9)
    let N2 = floor((month + 9)/12)
    let N3 = 1 + floor((year - 4*floor(year/4) + 2)/3)
    let dayOfYear = N1 - N2*N3 + day - 30

    for i in 0...1 {
        // Convert the longitude to hour value and calculate an approximate time
        let longitudeHour = longitude/15
        let t = dayOfYear + ((i == 0 ? 6.0 : 18.0) - longitudeHour)/24

        // Calculate the Sun's mean anomaly
        let M = 0.9856*t - 3.289

        // Calculate the Sun's true longitude
        var L = M + 1.916*sind(M) + 0.020*sind(2*M) + 282.634
        L %= 360

        // Calculate the Sun's right ascension
        var RA = atand(0.91764 * tand(L))
        RA %= 360
        let Lquadrant = (floor(L/90))*90
        let RAquadrant = (floor(RA/90))*90
        RA += Lquadrant - RAquadrant
        RA /= 15

        // Calculate the Sun's declination
        let sinDec = 0.39782*sind(L)
        let cosDec = cosd(asind(sinDec))

        // Calculate the Sun's local hour angle
        let cosH = (cosd(zenith.rawValue) - sinDec*sind(latitude))/(cosDec*cosd(latitude))
        if cosH > 1 { // The sun never rises on this location (on the specified date)
            sunriseTime = Double.infinity
            sunsetTime = -Double.infinity
        } else if cosH < -1 { // The sun never sets on this location (on the specified date)
            sunriseTime = -Double.infinity
            sunsetTime = Double.infinity
        } else {
            // Finish calculating H and convert into hours
            var H = ( i == 0 ? 360.0 : 0.0 ) + ( i == 0 ? -1.0 : 1.0 )*acosd(cosH)
            H /= 15

            // Calculate local mean time of rising/setting
            let T = H + RA - 0.06571*t - 6.622

            // Adjust back to UTC
            let UT = T - longitudeHour

            // Convert UT value to local time zone of latitude/longitude
            let localT = UT + localOffset

            if i == 0 { // Add 24 and modulo 24 to be sure that the results is between 0..<24
                sunriseTime = (localT + 24)%24
            } else {
                sunsetTime = (localT + 24)%24
            }
        }
    }
    return (sunriseTime, sunsetTime)
}


func sind(valueInDegrees: Double) -> Double {
    return sin(valueInDegrees*M_PI/180)
}

func cosd(valueInDegrees: Double) -> Double {
    return cos(valueInDegrees*M_PI/180)
}

func tand(valueInDegrees: Double) -> Double {
    return tan(valueInDegrees*M_PI/180)
}

func asind(valueInRadians: Double) -> Double {
    return asin(valueInRadians)*180/M_PI
}

func acosd(valueInRadians: Double) -> Double {
    return acos(valueInRadians)*180/M_PI
}

func atand(valueInRadians: Double) -> Double {
    return atan(valueInRadians)*180/M_PI
}

Ans这是我如何使用函数来确定是不是晚上：

let latitude = ...
let longitude = ...
let coordinate = CLLocationCoordinate2D(latitude: latitude, longitude: longitude)
let (sunriseHour, sunsetHour) = sunriseSunsetHoursForLocation(coordinate)
let componetns = NSCalendar.currentCalendar().components([.Hour, .Minute], fromDate: NSDate())
let currentHour = Double(componetns.hour) + Double(componetns.minute)/60
let isNight = currentHour < sunriseHour || currentHour > sunsetHour

Answer 1

我不确定为什么你用来获取偏移量的代码不能工作(我得到了同样的结果)。但有一个更简单的解决方案确实有效。只需使用secondsFromGMTForDate询问当地时区即可。如果日期间隔6个月，我会得到不同的结果：

let now = NSDate()
let future = NSCalendar.currentCalendar().dateByAddingUnit(NSCalendarUnit.Month, value: 6, toDate: now, options: NSCalendarOptions(rawValue: 0))!

let nowOffset = NSTimeZone.localTimeZone().secondsFromGMTForDate(now)/3600
let futureOffset = NSTimeZone.localTimeZone().secondsFromGMTForDate(future)/3600