我想从一年中的某天(1-366)和年份(例如 2011)转到 YYYYMMDD 格式的日期?
答案1
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
该函数将儒略零日视为前一年最后一天。
以下是基于 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" $1-01-01 +%Y%m%d; }
答案2
仅使用 Bash 无法完成,但如果你有 Perl:
use POSIX;
my ($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;
答案3
运行info 'Date input formats'
以查看允许哪些格式。
这YYYY-DDD 日期格式似乎不存在,并试图
$ date -d '2011-011'
date: invalid date `2011-011'
显示它不起作用,所以我认为njd
是正确的,最好的方法是使用除bash
和之外的外部工具date
。
如果您确实只想使用 bash 和基本命令行工具,您可以执行以下操作:
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
}
但这是一个相当缓慢的方法。
一种更快但更复杂的方法是尝试循环每个月,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
}
答案4
在 POSIX 终端上:
jul () { date -v$1y -v1m -v1d -v+$2d -v-1d "+%Y%m%d"; }
然后像调用
jul 2011 012
jul 2017 216
jul 2100 60