请高手帮帮忙，AIX4.3中，怎样用date实现取系统日期的前1天?

扁平足

大虾：
我在命令行输入‘date +%Y－%m－%d’可得到结果‘2008－09－25’，但是我想要当前日期的前1天或前5天的日期，请问需要加什么参数？
谢谢！

我老婆黑社会

这个要写shell了

我老婆黑社会

get_one_day_before_specified_date()
{
#get the command line input(date month & year)
day=$1
month=$2
year=$3

# if it is the first day of the month
if [ $day -eq 01 ]
then
# if it is the first month of the year
if [ $month -eq 01 ]
then
# make the month as 12
month=12

# deduct the year by one
year=`expr $year - 1`
else
# deduct the month by one
month=`expr $month - 1`
fi

# use cal command, discard blank lines,
# take last field of last line,
# first awk command is used to get the
# last useful line of the calendar cmd,
# second awk command is used to get the
# last field of this last useful line,
# NF is no. of fields,
# $NF is value of last field

day=`cal $month $year | awk 'NF != 0{ last = $0 }; END{ print last }' | awk '{ print $NF }'`
else
# deduct the day by one
day=`expr $day - 1`
fi

# echo $year$month$day
# year_2dig=`echo $year|awk '{print substr($0,3)}'`
echo $year$month$day
}

YEAR_T=`date +%Y`;
MONTH_T=`date +%m`;
DAY_T=`date +%d`;

YESTERDAY=`get_one_day_before_specified_date $DAY_T $MONTH_T $YEAR_T`;


DATE=$YESTERDAY

可以参考上面的

tyrone.dev

perl -e "use POSIX qw(strftime); print strftime '%Y%m%d' , localtime( time()-3600*24) "

扁平足

谢谢3、4楼的大侠们

CUaixer

改下时区就好了吖
TZ=TZ+1 这样子
不过最后要再改回来 要不影响其他东西
TZ=TZ