Linux下如何用date获取当前日期的前一天

今天遇到这个问题，从网上找到这两个合适的解答。

方法一：date命令的显示是与环境变量TZ有关的

$#看当前时区
$echo $TZ
CST-8
$#显示当前时间
$date
Mon Apr 2 15:48:36 CST 2002
$#改变当前时区，
TZ=CST+16;export TZ
$#显示当前时间（中间未改变系统时间，但date命令的显示已为昨天）
Mon Apr 1 15:48:33 CST 2002

=========================
方法二：一个详细的SCRIPT

#!/bin/sh

# ydate: A Bourne shell script that
# prints yestarday's date
# Output Form: Month Day Year
# From Focus on Unix: http://unix.about.com
# Set the current month day and year.
month=`date +%m`
day=`date +%d`
year=`date +%Y`

# Add 0 to month. This is a
# trick to make month an unpadded integer.
month=`expr $month + 0`

# Subtract one from the current day.
day=`expr $day - 1`

# If the day is 0 then determine the last
# day of the previous month.
if [ $day -eq 0 ]; then

# Find the preivous month.
month=`expr $month - 1`

# If the month is 0 then it is Dec 31 of
# the previous year.
if [ $month -eq 0 ]; then
month=12
day=31
year=`expr $year - 1`

# If the month is not zero we need to find
# the last day of the month.
else
case $month in
1|3|5|7|8|10|12) day=31;;
4|6|9|11) day=30;;
2)
if [ `expr $year % 4` -eq 0 ]; then
if [ `expr $year % 400` -eq 0 ]; then
day=29
elif [ `expr $year % 100` -eq 0 ]; then
day=28
else
day=29
fi
else
day=28
fi
;;
esac
fi
fi

# Print the month day and year.
echo $month $day $year
exit 0

============
如果是简单的只是想获得昨天的日期就很简单了，只要今天不是一号，呵呵
month=`date +%m`
day=`date +%d`
year=`date +%Y`
day=`expr $day - 1`

export DATEYMD=$year$month$day
echo $DATEYMD