How do I rename multiple files by removing characters in bash?
Solution 1:
A simple for loop with a bit of sed
will do the trick:
% touch xxxxx{foo,bar,baz}
% ls -l xxxxx{foo,bar,baz}
-rw-r--r-- 1 jamesog wheel 0 29 Dec 18:07 xxxxxbar
-rw-r--r-- 1 jamesog wheel 0 29 Dec 18:07 xxxxxbaz
-rw-r--r-- 1 jamesog wheel 0 29 Dec 18:07 xxxxxfoo
% for file in xxxxx*; do mv $file $(echo $file | sed -e 's/^.....//'); done
% ls -l foo bar baz
-rw-r--r-- 1 jamesog wheel 0 29 Dec 18:07 bar
-rw-r--r-- 1 jamesog wheel 0 29 Dec 18:07 baz
-rw-r--r-- 1 jamesog wheel 0 29 Dec 18:07 foo
The substitute regex in sed
says to match any five characters (.
means any character) at the start of the string (^
) and remove it.
Solution 2:
Bash has some amazing scripting possibilities. Here's one way:
for file in ??????*; do mv $file `echo $file | cut -c6-`; done
A handy way to test what it would do is to add an echo in front of the command:
for file in ??????*; do echo mv $file `echo $file | cut -c6-`; done
The six question marks ensure that you only attempt to do this to filenames longer than 5 characters.
Solution 3:
All great answers, thanks. This is what worked in my case:
rename 's/^.......//g' *
Solution 4:
You can use sed to do this
for file in * ; do mv $file $(echo $file |sed 's/^.\{5\}//g'); done