Question:
I have a directory full of files foo_num.txt. I'd like to rename all to num.txt (that is delete the "foo_" part). Can I do this in one line?
Solution:1
If you don't want to bother with for-loops in Bash, you might want to use the rename program:
rename "s/foo_//" *.txt The first argument is the Perl expression defining the string replacement rule. In this case: [s]ubstitute "foo_" with "".
The second argument filters the files you want to rename.
Solution:2
As your first part is separated by a _ I suggest you
rename 's/.*?_//' *.txt The ? means not greedy, therefore only the first occurrence of _ will be replaced.
Example
$ ls -laog total 4280 drwxrwxr-x 2 4329472 Aug 10 13:05 . drwx------ 55 20480 Aug 10 12:54 .. -rw-rw-r-- 1 0 Aug 10 13:05 foo_1_1.txt -rw-rw-r-- 1 0 Aug 10 13:05 foo_2_2.txt -rw-rw-r-- 1 0 Aug 10 13:05 foo_3_3.txt $ rename 's/.*?_//' *.txt $ ls -laog total 4280 drwxrwxr-x 2 4329472 Aug 10 13:06 . drwx------ 55 20480 Aug 10 12:54 .. -rw-rw-r-- 1 0 Aug 10 13:05 1_1.txt -rw-rw-r-- 1 0 Aug 10 13:05 2_2.txt -rw-rw-r-- 1 0 Aug 10 13:05 3_3.txt To replace all occurrences use
rename 's/.*_//' *.txt Example
$ ls -laog total 4280 drwxrwxr-x 2 4329472 Aug 10 13:08 . drwx------ 55 20480 Aug 10 12:54 .. -rw-rw-r-- 1 0 Aug 10 13:08 foo_1_1.txt -rw-rw-r-- 1 0 Aug 10 13:08 foo_2_2.txt -rw-rw-r-- 1 0 Aug 10 13:08 foo_3_3.txt $ rename 's/.*_//' *.txt $ ls -laog total 4280 drwxrwxr-x 2 4329472 Aug 10 13:09 . drwx------ 55 20480 Aug 10 12:54 .. -rw-rw-r-- 1 0 Aug 10 13:08 1.txt -rw-rw-r-- 1 0 Aug 10 13:08 2.txt -rw-rw-r-- 1 0 Aug 10 13:08 3.txt Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
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