无需 GNU 工具即可进行日期计算

Question 1

我的解决方案很简短，我之前已经用 C、Lua、Python、Perl、Awk、Ksh 和 Csh 编写过版本（下面的版本是 ksh。）您可以使用 ansi_dn、nd_isna 和 dow 回答一些具体问题：日期 YYYY MM DD，前后 100 天的日期是哪一天？给定一个日期，星期几？给定两个日期，中间相隔多少天？

#==========================================================================
# Calendar operations: Valid for dates in [1601-01-01, 3112-12-30].
# Adapted from a paper by B.E. Rodden, Mathematics Magazine, January, 1985.
#==========================================================================
# ANSI day number, given y=$1, m=$2, d=$3, starting from day number 1 ==
# (1601,1,1).  This is similar to COBOL's INTEGER-OF-DATE function.
function ansi_dn {
  integer y=$1 ys=$(($1-1601)) m=$2 d=$3 s dn1 isleap h
  ((isleap = y%4==0 && (y%100!=0 || y%400==0) ))
  ((dn1 = 365*ys + ys/4 - ys/100 + ys/400)) # first day number of given year.
  ((s = (214*(m-1) + 3)/7 + d - 1)) # standardized day number within year.
  ((h = -2*(!isleap && s>58) - (isleap && s>59) )) # correction to actual dn.
  print -- $((1 + dn1 + s + h))
}
# Inverse of ansi_dn, for adn=$1 in [1, 552245], with returned dates in
# [1601-01-01, 3112-12-30].  Similar to COBOL's DATE-OF-INTEGER function.
function nd_isna {
  integer y a s ys d0=$(($1-1)) dn1 isleap h
  typeset -Z2 m d
  ((ys = d0/365))
  ((365*ys + ys/4 - ys/100 + ys/400 >= $1)) && ((ys -= 1))
  ((dn1 = 365*ys + ys/4 - ys/100 + ys/400))
  ((y = 1601 + ys))
  ((a = d0 - dn1))
  ((isleap = y%4==0 && (y%100!=0 || y%400==0) ))
  ((h = -2*(!isleap && a>58) - (isleap && a>59) ))
  ((s = a - h))
  ((m = (7*s + 3)/214 + 1))
  ((d = s - (214*(m-1) + 3)/7 + 1))
  print -- $y $m $d
}
# Day of the week, given $1=ansi_dn.  0==Sunday, 6==Saturday.
function dow { print -- $(($1%7)); }

Answer

我的解决方案很简短，我之前已经用 C、Lua、Python、Perl、Awk、Ksh 和 Csh 编写过版本（下面的版本是 ksh。）您可以使用 ansi_dn、nd_isna 和 dow 回答一些具体问题：日期 YYYY MM DD，前后 100 天的日期是哪一天？给定一个日期，星期几？给定两个日期，中间相隔多少天？

#==========================================================================
# Calendar operations: Valid for dates in [1601-01-01, 3112-12-30].
# Adapted from a paper by B.E. Rodden, Mathematics Magazine, January, 1985.
#==========================================================================
# ANSI day number, given y=$1, m=$2, d=$3, starting from day number 1 ==
# (1601,1,1).  This is similar to COBOL's INTEGER-OF-DATE function.
function ansi_dn {
  integer y=$1 ys=$(($1-1601)) m=$2 d=$3 s dn1 isleap h
  ((isleap = y%4==0 && (y%100!=0 || y%400==0) ))
  ((dn1 = 365*ys + ys/4 - ys/100 + ys/400)) # first day number of given year.
  ((s = (214*(m-1) + 3)/7 + d - 1)) # standardized day number within year.
  ((h = -2*(!isleap && s>58) - (isleap && s>59) )) # correction to actual dn.
  print -- $((1 + dn1 + s + h))
}
# Inverse of ansi_dn, for adn=$1 in [1, 552245], with returned dates in
# [1601-01-01, 3112-12-30].  Similar to COBOL's DATE-OF-INTEGER function.
function nd_isna {
  integer y a s ys d0=$(($1-1)) dn1 isleap h
  typeset -Z2 m d
  ((ys = d0/365))
  ((365*ys + ys/4 - ys/100 + ys/400 >= $1)) && ((ys -= 1))
  ((dn1 = 365*ys + ys/4 - ys/100 + ys/400))
  ((y = 1601 + ys))
  ((a = d0 - dn1))
  ((isleap = y%4==0 && (y%100!=0 || y%400==0) ))
  ((h = -2*(!isleap && a>58) - (isleap && a>59) ))
  ((s = a - h))
  ((m = (7*s + 3)/214 + 1))
  ((d = s - (214*(m-1) + 3)/7 + 1))
  print -- $y $m $d
}
# Day of the week, given $1=ansi_dn.  0==Sunday, 6==Saturday.
function dow { print -- $(($1%7)); }

Question 2

过去我必须在 shell 脚本中通过暴力解析和计算来完成此操作。

在 shell 脚本中手动执行此操作非常容易出错且速度缓慢。您需要考虑每月的天数、闰年、时区。对于不同的区域设置和不同的语言，它会失败。

我修复了我的一个旧脚本，这可能需要针对不同的 shell 环境进行修改，但不应该有任何东西不能更改以在任何 shell 中工作。

#!/bin/sh 

datestring="Nov 28 20:27:19 2012 GMT"
today=`date`

function date2epoch {

month=$1
day=$2
time=$3
year=$4
zone=$5

# assume 365 day years
epochtime=$(( (year-1970) * 365 * 24 * 60 * 60 ))

# adjust for leap days
i=1970
while [[ $i -lt $4 ]]
do
    if [[ 0 -eq $(( i % 400 )) ]]
    then
        #echo $i is a leap year
        # divisible by 400 is a leap year
        epochtime=$((epochtime+24*60*60))
    elif [[ 0 -eq $(( i % 100 )) ]]
    then
        #echo $i is not a leap year
        epochtime=$epochtime
    elif [[ 0 -eq $(( i % 4 )) ]]
    then
        #echo $i is a leap year
        # divisible by 4 is a leap year
        epochtime=$((epochtime+24*60*60))
    #   epoch=`expr $epoch + 24 * 60 * 60`
    fi

    i=$((i+1))
done    

dayofyear=0
for imonth in Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
do
    if [[ $month == $imonth ]]
    then
        break
    fi

    case $imonth in
    'Feb')
        if [[ 0 -eq $(( year % 400 )) ]]
        then
            days=29
        elif [[ 0 -eq $(( year % 100 )) ]]
        then
            days=28
        elif [[ 0 -eq $(( year % 4 )) ]]
        then
            days=29
        fi
    ;;

    Jan|Mar|May|Jul|Aug|Oct|Dec) days=31 ;;
    *) days=30 ;;
    esac

    #echo $imonth has $days days
    dayofyear=$((dayofyear + days))
done
## add the day of the month 
dayofyear=$((dayofyear+day))
#echo $dayofyear

########## Add the day fo year (-1) to the epochtime
#(-1, since eg. Jan 1 is not 24 hours into Jan1 )

epochtime=$((epochtime + (dayofyear -1) * 24*60*60))
#echo $epochtime
################## hours, minutes, seconds
OFS=$IFS
IFS=":"
set -- $time
hours=$1
minutes=$2
seconds=$3
epochtime=$((epochtime + (hours * 60 * 60) + (minutes * 60) + seconds))
IFS=$OFS

################## Time zone

case $zone in
'GMT') zonenumber=0
break;;
'EST') zonenumber=-5
break;;
'EDT') zonenumber=-4
break;;
esac

epochtime=$((epochtime + zonenumber * 60 * 60 ))

echo $epochtime
}

result=`date2epoch $datestring`

echo $result

我可能在某个地方犯了错误，也许有更好的方法。如果您发现错误或更好的方法，请告诉我。

一旦你有了纪元时间，你就可以做一些有用的计算...尽管在没有 gnu utils 的情况下将其转换回日期...需要反向执行上述操作...

Answer