哪位大虾能帮忙将下面的sh程序写下注释啊,谢谢!
哪位大虾能帮忙将下面的sh程序写下注释啊,谢谢!
#!/bin/sh
SH_DIR=/u1/tmn/etc
export SH_DIR
DEL_PATH=/u1/tmn/bin/alcol/hw_omc/log
export DEL_PATH
. $SH_DIR/crontab.env
DT=`date "+%m%d "`
DA=`date "+%d "`
if [ "$DT " -eq "0101 " ]
then
CUR_YEAR=`date "+%y "`
DEL_YEAR=`expr $CUR_YEAR - 1`
DEL_DATE=`date "+200 "$DEL_YEAR "1231 "`
else
if [ "$DA " -eq "01 " ]
then
CUR_MON=`date "+%m "`
case $CUR_MON in
02,04,06,08,09 ) DEL_MON=`expr $CUR_MON - 1`
DEL_DATE=`date "+20%y "$DEL_MON "31 "`
;;
11 ) DEL_MON=`expr $CUR_MON - 1`
DEL_DATE=`date "+20%y "$DEL_MON "31 "`
;;
05,07,10 ) DEL_MON=`expr $CUR_MON - 1`
DEL_DATE=`date "+20%y "$DEL_MON "30 "`
;;
12 ) DEL_MON=`expr $CUR_MON - 1`
DEL_DATE=`date "+20%y "$DEL_MON "30 "`
;;
03 ) COU_YEAR=`date "+20%y "`
COU_MOD=`expr $COU_YEAR % 4`
if [ "$COU_MOD " -eq "0 " ]
then
DEL_DATE=`date "+20%y0229 "`
else
if [ "$COU_MOD " -ne "0 " ]
then
DEL_DATE=`date "+20%y0228 "`
fi
fi
esac
else
if [ "$DA " -lt "11 " ]
then
DEL_DAY=`expr $DA - 1`
DEL_DATE=`date "+20%y%m0 "$DEL_DAY " "`
else
if [ "$DA " -gt "10 " ]
then
DEL_DAY=`expr $DA - 1`
DEL_DATE=`date "+20%y%m "$DEL_DAY " "`
fi
fi
fi
fi
rm $DEL_PATH/HWOMC*. "$DEL_DATE "*
[解决办法]
就是找到当前日期的前一天，删除指定那个日期命名的文件。
那么多if,else,case什么的都是废话，一个date命令就可以得到当前日期的前一天。
[解决办法]
没有测试，你的sh脚本大概可以改写成这个样子：
#!/bin/sh
SH_DIR=/u1/tmn/etc
export SH_DIR
DEL_PATH=/u1/tmn/bin/alcol/hw_omc/log
export DEL_PATH
. $SH_DIR/crontab.env
DEL_DATE=`date -d "-1 day " +%Y%m%d`
rm $DEL_PATH/HWOMC*. "$DEL_DATE "*
linux通用。
