Using pure BASH:
s='10_I_am_here_001.jpg'
echo "${s#[0-9]*_}"
I_am_here_001.jpg
You can then write a simple for loop in that directory like this:
for s in *; do
f="${s#[0-9]*_}" && mv "$s" "$f"
done
Вопрос
I have a list of files like so :
10_I_am_here_001.jpg
20_I_am_here_003.jpg
30_I_am_here_008.jpg
40_I_am_here_004.jpg
50_I_am_here_009.jpg
60_I_am_here_002.jpg
70_I_am_here_005.jpg
80_I_am_here_006.jpg
How can I rename all the files in a directory, so that I can drop ^[0-9]+_
from the filename ?
Thank you
Решение
Using pure BASH:
s='10_I_am_here_001.jpg'
echo "${s#[0-9]*_}"
I_am_here_001.jpg
You can then write a simple for loop in that directory like this:
for s in *; do
f="${s#[0-9]*_}" && mv "$s" "$f"
done
Другие советы
Using rename
:
rename 's/^[0-9]+_//' *
Here's another bash idea based on files ending .jpg as shown above or whatever> VonBell
#!/bin/bash
ls *.jpg |\
while read FileName
do
NewName="`echo $FileName | cut -f2- -d "_"`"
mv $FileName $NewName
done
With bash extended globbing
shopt -s extglob
for f in *
do
[[ $f == +([0-9])_*.jpg ]] && mv "$f" "${f#+([0-9])_}"
done