我需要一个Shell脚本,从以下该文件名中获取
MMDDYYYY
信息。linuxbox.23566.MMDDYYYYHHMMSS.zip
使用 bash
字符串函数:
for file in *.zip; do
file="${file%.*}"
file="${file##*.}"
echo "${file:0:8}"
done
file="${file%.*}"
:去除扩展名并将新名称存储在file
变量中file="${file##*.}"
:从开头删除最长匹配,并将名称存储在file
变量中echo "${file:0:8}"
: echoes
剩下的前8个字符。$ ls
linuxbox.23566.MMDDYYYYHHMMSS.zip
$ for file in *; do file="${file%.*}"; file="${file##*.}"; echo "${file:0:8}"; done
MMDDYYYY
cut
命令:$ cut -d. -f3 <<< "linuxbox.23566.MMDDYYYYHHMMSS.zip" | cut -c-8
MMDDYYYY
$ cut -d. -f3 <<< "linuxbox.23566.MMDDYYYYHHMMSS.zip"
MMDDYYYYHHMMSS