通过删除字母从外壳重命名文件

Question

I have a lot of jpg files with letters and numbers in their names. I want to remove all letters, for example abc12d34efg.jpg becomes 1234.jpg . For the for loop I thought:

 for i in *.jpg; do mv "$i" ...

but I can't find a command for what I want.

Answer 1

With shell parameter expansion:

for fname in *.jpg; do mv "$fname" "${fname//[[:alpha:]]}jpg"; done

"${fname//[[:alpha:]]}" is a substitution of all occurrences of [[:alpha:]] (any letter) with nothing. Because this also removes the jpg , we have to add it again – the appended jpg does that.

Answer 2

您可以使用sed使用正则表达式将所有字母替换为空。

for i in *.jpg; do mv $i `echo $i | sed -e 's/[a-zA-Z]//g'`jpg; done

Answer 3

Here is a oneline solution that uses tr

 for f in *.jpg ; do n="$(echo ${f} | tr -d a-zA-Z )" ; mv "$f" "${n}jpg" ; done

With some formatting it would look like as

 for f in *.jpg ; do 
   n="$(echo ${f} | tr -d a-zA-Z )" 
   mv "$f" "${n}jpg"
 done

Here is what's happening:

First we remove all letters from the name using tr -d a-zA-Z . From abc12d34efg.jpg we get 1234. (with a dot at the end as . does not belong in the az and AZ intervals) and assign this value to variable $n . T

Then we can rename $f to ${n}jpg . That's it.

Update to delete both lower case and upper case letters use tr -d a-zA-Z , to delete only lower case letters use tr -d az instead.