UNIX中的模式匹配说明

Question

我有一个脚本如下：

for file in *.txt
do
    mv "$file" "${file%_*}.txt"
done

此脚本将文件CC_something_xyzgh_hh_ABCDEF.txt重命名为CC_something_xyzgh_hh.txt。但我不清楚到底是做什么的。

我在网上搜索过，但找不到任何能回答我问题的内容。

有人可以帮我吗？

Answer 1

你在这里基本上做的是shell parameter expansion。

${file%_*}从末尾删除扩展值的最短匹配模式。例如：

$ for file in *.txt; do echo "$file =====> ${file%_*}.txt"; done
CC_something_xyzgh_hh_ABCDEF.txt =====> CC_something_xyzgh_hh.txt

同样，要从最后删除最长匹配 ，您可以使用%%：

$ for file in *.txt; do echo "$file =====> ${file%%_*}.txt"; done 
CC_something_xyzgh_hh_ABCDEF.txt =====> CC.txt

要从开始的做类似的事情，您可以使用#（最短）和##（最长）

$ for file in *.txt; do echo "$file =====> ${file#*_}.txt"; done
CC_something_xyzgh_hh_ABCDEF.txt =====> something_xyzgh_hh_ABCDEF.txt.txt

$ for file in *.txt; do echo "$file =====> ${file##*_}.txt"; done
CC_something_xyzgh_hh_ABCDEF.txt =====> ABCDEF.txt.txt

Answer 2

${string%substring}

    Deletes shortest match of $substring from back of $string.

示例：

sat:~# file="CC_something_xyzgh_hh_ABCDEF.txt"
sat:~# echo ${file%_*}
CC_something_xyzgh_hh