How to 'rm' or 'mv' a range of files such as file01.txt, file02.txt...file85.txt

Question

Suppose I have a bunch of files named this way:

file01.txt file02.txt file03.txt ... file20.txt

and I want to execute a command on a range of these files.

I know if want to 'rm' from 'file05.txt' to 'file09.txt' I can do:

rm file0[5-9].txt

but how can I 'rm' a range from file08.txt to file13.txt? I tried this code

rm file[08-13].txt

and it doesn't work. I could run this command:

rm file0[89].txt file1[0-3].txt

but I'd like to know if I can do this with only one more refined argument to 'rm' if it is possible.

Answer 1

You need to use brace expansion of bash:

rm file{08..13}.txt

This will remove files file08 to file13, will show an error message if any file is missing.

Set the range to meet your need.

---

The problem with the globbing operator [] is that it treats each character/digit inside it as a single token and so only supports range declaration using single digits.

If you insist on using [], you can use this rather ugly pattern to match file08 to file13:

rm file0[8-9].txt file1[0-3].txt

Answer 2

Some shells, such as Ubuntu's default /bin/sh ( which is dash ) or mksh don't have brace expansion, or in case of ksh - brace expansion can't use padded zeroes :

$ ksh -c 'echo {05..13}'
5 6 7 8 9 10 11 12 13

In such cases, we can make use of printf to format the number portion of the filename, and use a while loop to implement c-like for loop behavior (note to replace echo with rm or whatever you want):

$ i=5; while [ "$i" -le 10 ]; do num=$(printf "%.2d" "$i" ); echo "file$num.txt";i=$(($i+1)); done
file05.txt
file06.txt
file07.txt
file08.txt
file09.txt
file10.txt

And this is fairly portable - works with dash, ksh, mksh, also bash. In the case of ksh and bash we can also use c-style for loop syntax (but not in case ofmkshordash`):

$ ksh -c 'for((i=5;i<11;i++)); do num=$(printf "%.2d" "$i" ); echo "file$num.txt";done'             
file05.txt
file06.txt
file07.txt
file08.txt
file09.txt
file10.txt

Note that in bash's case, printf supports printing to variable, and thus we could do printf -v num "%.2d" "$i" instead of using command substitution.