Javascript或Python如何判断是白天还是夜晚？

Question

知道我如何根据用户的时间和位置判断当前是晚上/白天还是日出/黎明？

我没有发现任何有用的东西可以在客户端或后端使用。在

让它变得棘手的是，小时并不一定定义它是否是黑夜和白天，这在很大程度上取决于年、月、日、时和地理坐标。在

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为了澄清。。。模仿这样的东西。在

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一种近似的方法也非常有用。

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希望有人能帮忙！在

Answer 1

你可以像我一样，使用这个公共域Sun.py模块来计算太阳相对于地球位置的位置。它很旧，但多年来对我很管用。我对它做了一些表面上的修改，使之更能适应Python2.7的最新版本，比如使其中的几个类成为新样式，但在大多数情况下它没有改变。在

这是我创建的一个模块，名为日出集.py，它展示了如何使用它来计算给定地理坐标和时区的特定位置的日出和日落时间。引用的timezone模块是datetime模块文档中描述的^{}抽象基类的实现。在

# -*- coding: iso-8859-1 -*-
import datetime
import timezone  # concrete tzinfo subclass based on the Python docs
import math
from Sun import Sun

__all__ = ['getsuninfo', 'Place']

class Place(object):
    def __init__(self, name, coords, tz=timezone.Pacific):
        self.name = name        # string
        self.coords = coords    # tuple (E/W long, N/S lat)
        self.tz = tz            # tzinfo constant

def _hoursmins(hours):
    """Convert floating point decimal time in hours to integer hrs,mins"""
    frac,h = math.modf(hours)
    m = round(frac*60, 0)
    if m == 60: # rounded up to next hour
        h += 1; m = 0
    return int(h),int(m)

def _ymd(date):
    """Return y,m,d from datetime object as tuple"""
    return date.timetuple()[:3]

def getsuninfo(location, date=None):
    """Return local datetime of sunrise, sunset, and length of day in hrs,mins)"""
    if date == None:
        querydate = datetime.date.today()
    else: # date given should be datetime instance
        querydate = date

    args = _ymd(querydate) + location.coords
    utcrise, utcset = Sun().sunRiseSet(*args)
    daylength = Sun().dayLength(*args)
    hrs,mins = _hoursmins(daylength)

    risehour, risemin = _hoursmins(utcrise)
    sethour, setmin   = _hoursmins(utcset)

    # convert times to timedelta values (ie from midnight utc of the date)
    midnight = datetime.datetime(tzinfo=timezone.utc, *_ymd(querydate))
    deltarise = datetime.timedelta(hours=risehour, minutes=risemin)
    utcdatetimerise = midnight+deltarise
    deltaset = datetime.timedelta(hours=sethour, minutes=setmin)
    utcdatetimeset  = midnight+deltaset

    # convert results from UTC time to local time of location
    localrise = utcdatetimerise.astimezone(location.tz)
    localset  = utcdatetimeset.astimezone(location.tz)

    return localrise, localset, hrs, mins

if __name__ == "__main__":
    import datetime, timezone

    def unittest(location, testdate):
        risetime, settime, hrs, mins = getsuninfo(location, testdate)

        print "Location:", location.name
        print "Date:", testdate.strftime("%a %x")
        print risetime.strftime("Sunrise %I:%M %p"), settime.strftime("- Sunset %I:%M %p (%Z)")
        print "daylight: %d:%02d" % (hrs,mins)
        print

    place = Place("My House", (-121.990278, 47.204444), timezone.Pacific)

    # test dates just before and after DST transitions
    print "pre 2007"
    print "========="
    unittest(place, datetime.date(2006, 4, 1))
    unittest(place, datetime.date(2006, 4, 2))
    unittest(place, datetime.date(2006, 10, 28))
    unittest(place, datetime.date(2006, 10, 29))

    print "2007"
    print "========="
    unittest(place, datetime.date(2007, 3, 10))
    unittest(place, datetime.date(2007, 3, 11))
    unittest(place, datetime.date(2007, 11, 3))
    unittest(place, datetime.date(2007, 11, 4))

Answer 2

美国海军天文台对计算日出和日落的算法进行了简要说明，可在以下网址获得：

http://edwilliams.org/sunrise_sunset_algorithm.htm

除了提供日期和位置，您还需要选择一个天顶角（在这个角度上太阳将被视为“升起”或“落下”）—链接的页面有几个选项。在

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更新

因为下面的文字已经链接不上了。请注意，包含的公式是伪代码形式，而不是JavaScript。在

Source:
    Almanac for Computers, 1990
    published by Nautical Almanac Office
    United States Naval Observatory
    Washington, DC 20392

Inputs:
    day, month, year:      date of sunrise/sunset
    latitude, longitude:   location for sunrise/sunset
    zenith:                Sun's zenith for sunrise/sunset
      offical      = 90 degrees 50'
      civil        = 96 degrees
      nautical     = 102 degrees
      astronomical = 108 degrees

    NOTE: longitude is positive for East and negative for West
        NOTE: the algorithm assumes the use of a calculator with the
        trig functions in "degree" (rather than "radian") mode. Most
        programming languages assume radian arguments, requiring back
        and forth convertions. The factor is 180/pi. So, for instance,
        the equation RA = atan(0.91764 * tan(L)) would be coded as RA
        = (180/pi)*atan(0.91764 * tan((pi/180)*L)) to give a degree
        answer with a degree input for L.


1. first calculate the day of the year

    N1 = floor(275 * month / 9)
    N2 = floor((month + 9) / 12)
    N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))
    N = N1 - (N2 * N3) + day - 30

2. convert the longitude to hour value and calculate an approximate time

    lngHour = longitude / 15

    if rising time is desired:
      t = N + ((6 - lngHour) / 24)
    if setting time is desired:
      t = N + ((18 - lngHour) / 24)

3. calculate the Sun's mean anomaly

    M = (0.9856 * t) - 3.289

4. calculate the Sun's true longitude

    L = M + (1.916 * sin(M)) + (0.020 * sin(2 * M)) + 282.634
    NOTE: L potentially needs to be adjusted into the range [0,360) by adding/subtracting 360

5a. calculate the Sun's right ascension

    RA = atan(0.91764 * tan(L))
    NOTE: RA potentially needs to be adjusted into the range [0,360) by adding/subtracting 360

5b. right ascension value needs to be in the same quadrant as L

    Lquadrant  = (floor( L/90)) * 90
    RAquadrant = (floor(RA/90)) * 90
    RA = RA + (Lquadrant - RAquadrant)

5c. right ascension value needs to be converted into hours

    RA = RA / 15

6. calculate the Sun's declination

    sinDec = 0.39782 * sin(L)
    cosDec = cos(asin(sinDec))

7a. calculate the Sun's local hour angle

    cosH = (cos(zenith) - (sinDec * sin(latitude))) / (cosDec * cos(latitude))

    if (cosH >  1) 
      the sun never rises on this location (on the specified date)
    if (cosH < -1)
      the sun never sets on this location (on the specified date)

7b. finish calculating H and convert into hours

    if if rising time is desired:
      H = 360 - acos(cosH)
    if setting time is desired:
      H = acos(cosH)

    H = H / 15

8. calculate local mean time of rising/setting

    T = H + RA - (0.06571 * t) - 6.622

9. adjust back to UTC

    UT = T - lngHour
    NOTE: UT potentially needs to be adjusted into the range [0,24) by adding/subtracting 24

10. convert UT value to local time zone of latitude/longitude

    localT = UT + localOffset

Answer 3

PyEphem可以用来计算到下一次日出和日落的时间。在a blog post I found和{a3}的基础上，您的问题可以按如下方式解决。假设我是您的用户，我的位置是Oldenburg (Oldb), Germany。在

import ephem

user = ephem.Observer()
user.lat = '53.143889'    # See wikipedia.org/Oldenburg
user.lon = '8.213889'     # See wikipedia.org/Oldenburg
user.elevation = 4        # See wikipedia.org/Oldenburg
user.temp = 20            # current air temperature gathered manually
user.pressure = 1019.5    # current air pressure gathered manually

next_sunrise_datetime = user.next_rising(ephem.Sun()).datetime()
next_sunset_datetime = user.next_setting(ephem.Sun()).datetime()

# If it is daytime, we will see a sunset sooner than a sunrise.
it_is_day = next_sunset_datetime < next_sunrise_datetime
print("It's day." if it_is_day else "It's night.")

# If it is nighttime, we will see a sunrise sooner than a sunset.
it_is_night = next_sunrise_datetime < next_sunset_datetime
print("It's night." if it_is_night else "It's day.")

注释

• 由于某些原因，lat和{}需要是字符串，但是如果它们是float，则ephem不会抱怨。在
• 为了获得最佳效果，您可能需要获得当前的空气温度和空气压力。在

先决条件

这至少适用于Python2.7（使用pip-2.7 install pyephem）和Python3.2（使用pip-3.2 install ephem）。在

确保在系统上运行网络时间协议客户端。E、 g.在Debian Linux上：

^{pr2}$

请确保在您的系统上设置了正确的时区。E、 g.在Debian Linux上：

$ sudo dpkg-reconfigure tzdata