如何在Bash中打印当前时间前一天的日期?


当前回答

#!/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

其他回答

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

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
yesterday=`date -d "-1 day" %F`

将昨天的日期(YYYY-MM-DD格式)放入变量$yesterday。

如果你有BSD (OSX)日期,你可以这样做:

date -j -v-1d
Wed Dec 14 15:34:14 CET 2011

或者如果你想在任意日期上进行日期计算:

date -j -v-1d -f "%Y-%m-%d" "2011-09-01" "+%Y-%m-%d"
2011-08-31
date -d "yesterday" '+%Y-%m-%d'

or

date=$(date -d "yesterday" '+%Y-%m-%d')
echo $date