AIX环境下计算昨天日期的shell两例

1、 改变时区的方法

cat yesterday.sh

 aaa=`echo $TZ|sed 's/.*/(../)//1/'` aaa=`expr $aaa + 24` eval aaa=`echo $TZ|sed 's/..$/+$aaa/'` TZ=$aaa export TZ yy=`date +%Y` mm=`date +%m` dd=`date +%d` echo $yy$mm$dd

2、 遍历所有情况类似函数的方法

cat yesterday.sh

 #!/bin/bash # Var Declare #!/bin/sh # ydate: A Bourne shell script that # prints yestarday's date # Output Form: Month Day Year # 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 $year % 400` -eq 0 ]; then day=29 fi else day=28 fi ;; esac fi fi echo yesterday:$year-${month}-${day}

说明：方法1较简洁明了，代码清晰；方法2是一般写程序的方法。

3、 2中方法的进一步更新（支持20110101格式，即，年月固定用两位数）

cat yesterday.sh

#!/bin/bash # Var Declare #!/bin/sh # ydate: A Bourne shell script that # prints yestarday's date # Output Form: Month Day Year # 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 fi else day=28 fi ;; esac fi fi case $day in 1|2|3|4|5|6|7|8|9) day='0'$day esac case $month in 1|2|3|4|5|6|7|8|9) month='0'$month esac echo $year$month$day

说明：经过小小的改变，对实际应用提供了极大的方便。