#!/bin/bash
OFFSET=1;
eval `date "+day=%d; month=%m; year=%Y"`
# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=`expr $day - $OFFSET`
if [ $day -le 0 ] ;then
month=`expr $month - 1`
if [ $month -eq 0 ] ;then
year=`expr $year - 1`
month=12
fi
set `cal $month $year`
xday=${$#}
day=`expr $xday + $day`
fi
echo $year-$month-$day

#!/bin/bash
OFFSET=1;
eval `date "+day=%d; month=%m; year=%Y"`
# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=`expr $day - $OFFSET`
if [ $day -le 0 ] ;then
month=`expr $month - 1`
if [ $month -eq 0 ] ;then
year=`expr $year - 1`
month=12
fi
set `cal $month $year`
xday=${$#}
day=`expr $xday + $day`
fi
echo $year-$month-$day

date +%Y:%m:%d|awk -vFS=":" -vOFS=":" '{$3=$3-1;print}'
2009:11:9

不是很性感，但可以胜任:

perl -e 'my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time - 86400);$year += 1900; $mon+= 1; printf ("YESTERDAY: %04d%02d%02d \n", $year, $mon, $mday)'

由“马丁·克莱顿”的答案形成。

如果你有GNU日期，我理解正确

$ date +%Y:%m:%d -d "yesterday"
2009:11:09

or

$ date +%Y:%m:%d -d "1 day ago"
2009:11:09

MAC OSX

昨天的日期:

date -v-1d +%F

其中1d定义今天减去1天。同样的,

date -v-1w +%F -表示前一周的日期

date -v-1m +%F -表示上个月的日期

如果你有gnu日期，

date --date="1 day ago"

更多信息:https://www.cyberciti.biz/tips/linux-unix-get-yesterdays-tomorrows-date.html