怎么把一年中的年月日转换成日期YYYYMMDD？

我要从年份（1-366）和年份（例如2011）转到格式为YYYYMMDD的日期？

这个Bash函数在基于GNU的系统上对我有用：

jul () { date -d "$1-01-01 +$2 days -1 day" "+%Y%m%d"; }

一些例子：

$ y=2011; od=0; for d in {-4..4} 59 60 {364..366} 425 426; do (( d > od + 1)) && echo; printf "%3s " $d; jul $y $d; od=$d; done
 -4 20101227
 -3 20101228
 -2 20101229
 -1 20101230
  0 20101231
  1 20110101
  2 20110102
  3 20110103
  4 20110104

 59 20110228
 60 20110301

364 20111230
365 20111231
366 20120101

425 20120229
426 20120301

此功能将儒略日零视为上一年的最后一天。

这是基于UNIX的系统（例如macOS）的bash函数：

jul () { (( $2 >=0 )) && local pre=+; date -v$pre$2d -v-1d -j -f "%Y-%m-%d" $1-01-01 +%Y%m%d; }

不能仅用Bash来完成，但是如果您有Perl：

use POSIX;

my ($jday, $year) = (100, 2011);

# Unix time in seconds since Jan 1st 1970
my $time = mktime(0,0,0, $jday, 0, $year-1900);

# same thing as a list that we can use for date/time formatting
my @tm = localtime $time;

my $yyyymmdd = strftime "%Y%m%d", @tm;

运行info 'Date input formats'查看允许的格式。

该YYYY-DDD日期格式似乎并不存在，并试图

$ date -d '2011-011'
date: invalid date `2011-011'

显示它不起作用，因此我认为njd是正确的，最好的方法是使用bashand 以外的外部工具date。

如果您真的只想使用bash和基本命令行工具，则可以执行以下操作：

julian_date_to_yyyymmdd()
{
    date=$1    # assume all dates are in YYYYMMM format
    year=${date%???}
    jday=${date#$year}
    for m in `seq 1 12`; do
        for d in `seq 1 31`; do
            yyyymmdd=$(printf "%d%02d%02d" $year $m $d)
            j=$(date +"%j" -d "$yyyymmdd" 2>/dev/null)
            if test "$jday" = "$j"; then
                echo "$yyyymmdd"
                return 0
            fi
        done
    done
    echo "Invalid date" >&2
    return 1
}

但这是一个很慢的方法。

一种更快但更复杂的方法尝试遍历每个月，找到该月的最后一天，然后查看儒略日是否在该范围内。

# year_month_day_to_jday <year> <month> <day> => <jday>
# returns 0 if date is valid, non-zero otherwise
# year_month_day_to_jday 2011 2 1 => 32
# year_month_day_to_jday 2011 1 32 => error
year_month_day_to_jday()
{
    # XXX use local or typeset if your shell supports it
    _s=$(printf "%d%02d%02d" "$1" "$2" "$3")
    date +"%j" -d "$_s"
}

# last_day_of_month_jday <year> <month>
# last_day_of_month_jday 2011 2 => 59
last_day_of_month_jday()
{
    # XXX use local or typeset if you have it
    _year=$1
    _month=$2
    _day=31

    # GNU date exits with 0 if day is valid, non-0 if invalid
    # try counting down from 31 until we find the first valid date
    while test $_day -gt 0; do
        if _jday=$(year_month_day_to_jday $_year $_month $_day 2>/dev/null); then
            echo "$_jday"
            return 0
        fi
        _day=$((_day - 1))
    done
    echo "Invalid date" >&2
    return 1
}

# first_day_of_month_jday <year> <month>
# first_day_of_month_jday 2011 2 => 32
first_day_of_month_jday()
{
    # XXX use local or typeset if you have it
    _year=$1
    _month=$2
    _day=1

    if _jday=$(year_month_day_to_jday $_year $_month 1); then
        echo "$_jday"
        return 0
    else
        echo "Invalid date" >&2
        return 1
    fi
}

# julian_date_to_yyyymmdd <julian day> <4-digit year>
# e.g. julian_date_to_yyyymmdd 32 2011 => 20110201
julian_date_to_yyyymmdd()
{
    jday=$1
    year=$2

    for m in $(seq 1 12); do
        endjday=$(last_day_of_month_jday $year $m)
        if test $jday -le $endjday; then
            startjday=$(first_day_of_month_jday $year $m)
            d=$((jday - startjday + 1))
            printf "%d%02d%02d\n" $year $m $d
            return 0
        fi
    done
    echo "Invalid date" >&2
    return 1
}

如果年份（1-366）是149，年份是2014，

$ date -d "148 days 2014-01-01" +"%Y%m%d"
20140529

请确保输入年份-1的值。

我在bash中的解决方案

from_year=2013
from_day=362

to_year=2014
to_day=5

now=`date +"%Y/%m/%d" -d "$from_year/01/01 + $from_day days - 2 day"`
end=`date +"%Y/%m/%d" -d "$to_year/01/01 + $to_day days - 1 day"`


while [ "$now" != "$end" ] ; 
do
    now=`date +"%Y/%m/%d" -d "$now + 1 day"`;
    echo "$now";
    calc_day=`date -d "$now" +%G'.'%j`
    echo $calc_day
done

在POSIX终端上：

jul () { date -v$1y -v1m -v1d -v+$2d -v-1d "+%Y%m%d"; }

然后像

jul 2011 012
jul 2017 216
jul 2100 60

我意识到这是很久以前问过的，但是我看到的答案都不是我认为的纯BASH，因为它们都使用GNU date。我以为我会回答...但是我会提前警告您，答案不是很优雅，也不是超过100行。可以简化它，但是我想让其他人轻松查看它的作用。

这里的主要“技巧”是通过%将年份的MODULUS 除以4，然后将每个月中的天数加起来（如果需要的话，加上2月的额外天数）来确定一年是否为a年。，使用简单的值表。

请随意发表评论并提出改进方法的建议，因为我主要是在这里学习更多自己的知识，并且我认为自己充其量是BASH的新手。我已尽我所能，做到了随身携带，这在我看来意味着一些折衷。

关于代码...我希望它是不言自明的。

#!/bin/sh

# Given a julian day of the year and a year as ddd yyyy, return
# the values converted to yyyymmdd format using ONLY bash:

ddd=$1
yyyy=$2
if [ "$ddd" = "" ] || [ "$yyyy" = "" ]
then
  echo ""
  echo "   Usage: <command> 123 2016"
  echo ""
  echo " A valid julian day from 1 to 366 is required as the FIRST"
  echo " parameter after the command, and a valid 4-digit year from"
  echo " 1901 to 2099 is required as the SECOND. The command and each"
  echo " of the parameters must be separated from each other with a space."
  echo " Please try again."
  exit
fi
leap_yr=$(( yyyy % 4 ))
if [ $leap_yr -ne 0 ]
then
  leap_yr=0
else
  leap_yr=1
fi
last_doy=$(( leap_yr + 365 ))
while [ "$valid" != "TRUE" ]
do
  if [ 0 -lt "$ddd" ] && [ "$last_doy" -ge "$ddd" ]
  then
    valid="TRUE"
  else
    echo "   $ddd is an invalid julian day for the year given."
    echo "   Please try again with a number from 1 to $last_doy."
    exit    
  fi
done
valid=
while [ "$valid" != "TRUE" ]
do
  if [ 1901 -le "$yyyy" ] && [ 2099 -ge "$yyyy" ]
  then
    valid="TRUE"
  else
    echo "   $yyyy is an invalid year for this script."
    echo "   Please try again with a number from 1901 to 2099."
    exit    
  fi
done
if [ "$leap_yr" -eq 1 ]
then
  jan=31  feb=60  mar=91  apr=121 may=152 jun=182
  jul=213 aug=244 sep=274 oct=305 nov=335
else
  jan=31  feb=59  mar=90  apr=120 may=151 jun=181
  jul=212 aug=243 sep=273 oct=304 nov=334
fi
if [ "$ddd" -gt $nov ]
then
  mm="12"
  dd=$(( ddd - nov ))
elif [ "$ddd" -gt $oct ]
then
  mm="11"
  dd=$(( ddd - oct ))
elif [ "$ddd" -gt $sep ]
then
  mm="10"
  dd=$(( ddd - sep ))
elif [ "$ddd" -gt $aug ]
then
  mm="09"
  dd=$(( ddd - aug ))
elif [ "$ddd" -gt $jul ]
then
  mm="08"
  dd=$(( ddd - jul ))
elif [ "$ddd" -gt $jun ]
then
  mm="07"
  dd=$(( ddd - jun ))
elif [ "$ddd" -gt $may ]
then
  mm="06"
  dd=$(( ddd - may ))
elif [ "$ddd" -gt $apr ]
then
  mm="05"
  dd=$(( ddd - apr ))
elif [ "$ddd" -gt $mar ]
then
  mm="04"
  dd=$(( ddd - mar ))
elif [ "$ddd" -gt $feb ]
then
  mm="03"
  dd=$(( ddd - feb ))
elif [ "$ddd" -gt $jan ]
then
  mm="02"
  dd=$(( ddd - jan ))
else
  mm="01"
  dd="$ddd"
fi
if [ ${#dd} -eq 1 ]
then
  dd="0$dd"
fi
if [ ${#yyyy} -lt 4 ]
then
  until [ ${#yyyy} -eq 4 ]
  do
    yyyy="0$yyyy"
  done
fi
printf '\n   %s%s%s\n\n' "$yyyy" "$mm" "$dd"

OLD_JULIAN_VAR=$(date -u -d 1840-12-31 +%s)

TODAY_DATE=`date --date="$odate" +"%Y-%m-%d"`
TODAY_DATE_VAR=`date -u -d "$TODAY_DATE" +"%s"`
export JULIAN_DATE=$((((TODAY_DATE_VAR - OLD_JULIAN_VAR))/86400))
echo $JULIAN_DATE

在数学上表示如下

[(date in sec)-(1840-12-31 in sec)]/(sec in a day 86400)