如何在Bash中打印当前时间前一天的日期?
当前回答
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
其他回答
好吧,这是一个晚的答案,但这似乎是有效的!
YESTERDAY=`TZ=GMT+24 date +%d-%m-%Y`;
echo $YESTERDAY;
#!/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
试试下面的代码,它也处理DST部分。
if [ $(date +%w) -eq $(date -u +%w) ]; then
tz=$(( 10#$gmthour - 10#$localhour ))
else
tz=$(( 24 - 10#$gmthour + 10#$localhour ))
fi
echo $tz
myTime=`TZ=GMT+$tz date +'%Y%m%d'`
何赛义工
date --date='-1 day'
date +%Y:%m:%d|awk -vFS=":" -vOFS=":" '{$3=$3-1;print}'
2009:11:9