学步园

1、 改变时区的方法

cat yesterday.sh

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

2、 遍历所有情况类似函数的方法

cat yesterday.sh

 #!/bin/bash<br /> # Var Declare<br /> #!/bin/sh<br /> # ydate: A Bourne shell script that<br /> # prints yestarday's date<br /> # Output Form: Month Day Year</p> <p># Set the current month day and year.<br /> month=`date +%m`<br /> day=`date +%d`<br /> year=`date +%Y`</p> <p># Add 0 to month. This is a<br /> # trick to make month an unpadded integer.<br /> month=`expr $month + 0`</p> <p># Subtract one from the current day.<br /> day=`expr $day - 1`</p> <p># If the day is 0 then determine the last<br /> # day of the previous month.<br /> if [ $day -eq 0 ]; then</p> <p># Find the preivous month.<br /> month=`expr $month - 1`</p> <p># If the month is 0 then it is Dec 31 of<br /> # the previous year.<br /> if [ $month -eq 0 ]; then<br /> month=12<br /> day=31<br /> year=`expr $year - 1`</p> <p># If the month is not zero we need to find<br /> # the last day of the month.<br /> else<br /> case $month in<br /> 1|3|5|7|8|10|12) day=31;;<br /> 4|6|9|11) day=30;;<br /> 2)<br /> if [ `expr $year % 4` -eq 0 ]; then<br /> if [ `expr $year % 400` -eq 0 ]; then<br /> day=29<br /> fi<br /> else<br /> day=28<br /> fi<br /> ;;<br /> esac<br /> fi<br /> fi</p> <p>echo yesterday:$year-${month}-${day}<br />

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<br /> # Var Declare<br /> #!/bin/sh<br /> # ydate: A Bourne shell script that<br /> # prints yestarday's date<br /> # Output Form: Month Day Year</p> <p># Set the current month day and year.<br /> month=`date +%m`<br /> day=`date +%d`<br /> year=`date +%Y`</p> <p># Add 0 to month. This is a<br /> # trick to make month an unpadded integer.<br /> month=`expr $month + 0`</p> <p># Subtract one from the current day.<br /> day=`expr $day - 1`</p> <p># If the day is 0 then determine the last<br /> # day of the previous month.<br /> if [ $day -eq 0 ]; then</p> <p># Find the preivous month.<br /> month=`expr $month - 1`</p> <p># If the month is 0 then it is Dec 31 of<br /> # the previous year.<br /> if [ $month -eq 0 ]; then<br /> month=12<br /> day=31<br /> year=`expr $year - 1`</p> <p># If the month is not zero we need to find<br /> # the last day of the month.<br /> else<br /> case $month in<br /> 1|3|5|7|8|10|12) day=31;;<br /> 4|6|9|11) day=30;;<br /> 2)<br /> if [ `expr $year % 4` -eq 0 ]; then<br /> if [ `expr $year % 400` -eq 0 ]; then<br /> day=29<br /> fi<br /> else<br /> day=28<br /> fi<br /> ;;<br /> esac<br /> fi<br /> fi</p> <p>case $day<br /> in 1|2|3|4|5|6|7|8|9) day='0'$day<br /> esac<br /> case $month<br /> in 1|2|3|4|5|6|7|8|9) month='0'$month<br /> esac<br /> echo $year$month$day<br />

说明：经过小小的改变，对实际应用提供了极大的方便。