rename multiple files with regular-expression

Question

Inside a folder I have a file structure, the files have all a name whose format

name-randomstring.extension

eg

./dir1/aaa-5h34jk5hk.js
./dir2/bbb-5yh45uh9h.css
./dir3/ccc-uiero6tio.js
./dir3/ddd-7y8fygfre.css
. . .

with a bash script I would like to rename them recursively; so to eliminate the -randomstring from every file

./dir1/aaa.js
./dir2/bbb.css
./dir3/ccc.js
./dir3/ddd.css
. . .

@Tom: It's no duplicate. The link should only show that it is possible to use sed's syntax with rename command. — Cyrus, Sep 16 '17 at 09:36
You mean because the file pattern is not exactly the same, it's not a duplicate? I disagree. — Tom Fenech, Sep 16 '17 at 09:38
Do you have the other directories inside that directory? I mean do you need renaming recursively? — Shakiba Moshiri, Sep 16 '17 at 09:50
yes need renaming recursively, with find command and piping? — Simone Bonelli, Sep 16 '17 at 10:02
Possible duplicate of https://stackoverflow.com/q/16541582/2088135 in that case. — Tom Fenech, Sep 16 '17 at 10:03
@Tom thanks, this question helps me to set the recursive command with "find", but how to rename the single file? $ find . -exec rename ????? '{}' \; — Simone Bonelli, Sep 16 '17 at 10:09

Shakiba Moshiri · Answer 1 · 2017-09-16T10:41:34.697

if you do not mind using Perl then you can do anything you like.

for example for renaming all files with this structure:

dir > find .
.
./00_file.txt
./01_file.txt
./02_file.txt
./dir1
./dir1/00_file.txt
./dir1/01_file.txt
./dir1/02_file.txt
./dir2
./dir2/00_file.txt
./dir2/01_file.txt
./dir2/02_file.txt
./dir3
./dir3/00_file.txt
./dir3/01_file.txt
./dir3/02_file.txt   
./find.log

then save your list in a file like: find.log now you can use Perl

dir > perl -lne '-f && ($old=$_) && s/file/ABCD/g && print "$old => $_"' find.log
./00_file.txt => ./00_ABCD.txt
./01_file.txt => ./01_ABCD.txt
./02_file.txt => ./02_ABCD.txt
./dir1/00_file.txt => ./dir1/00_ABCD.txt
./dir1/01_file.txt => ./dir1/01_ABCD.txt
./dir1/02_file.txt => ./dir1/02_ABCD.txt
./dir2/00_file.txt => ./dir2/00_ABCD.txt
./dir2/01_file.txt => ./dir2/01_ABCD.txt
./dir2/02_file.txt => ./dir2/02_ABCD.txt
./dir3/00_file.txt => ./dir3/00_ABCD.txt
./dir3/01_file.txt => ./dir3/01_ABCD.txt
./dir3/02_file.txt => ./dir3/02_ABCD.txt

how it works?

read the file
let only file; not directory or other stuffs with: -f
then save the old name: ($old=$_)
then doing substitution with: s///g operator
finally rename the old file to new one that is $_

NOTE I used print and you should use rename like this:

dir > perl -lne '-f && ($old=$_) && s/file/ABCD/g && rename $old, $_' find.log
dir > find .
.
./00_ABCD.txt
./01_ABCD.txt
./02_ABCD.txt
./dir1
./dir1/00_ABCD.txt
./dir1/01_ABCD.txt
./dir1/02_ABCD.txt
./dir2
./dir2/00_ABCD.txt
./dir2/01_ABCD.txt
./dir2/02_ABCD.txt
./dir3
./dir3/00_ABCD.txt
./dir3/01_ABCD.txt
./dir3/02_ABCD.txt
./find.log

NOTE Since the find returns list of all files and sub-directories, with this technique Perl renames all the files! Thus first use print then rename. And your pattern can be: -.*(?=\.). In fact:

s/-.*(?=\.)//g

score 0 · Answer 2 · answered Sep 16 '17 at 10:39

You can do this with a native bash loop:

shopt -s globstar # enable ** for recursive expansion
for file in **/*{j,cs}; do
    # remove everything up to the last . to get extension
    ext=${file##*.}

    # remove everything after the last - and concat with extension
    new_name=${file%-*}$ext 

    # -- prevents weird filenames being interpreted as options
    mv -- "$file" "$new_name" 
done
shopt -u globstar # disable ** if you don't want it anymore e.g. in a script

rename multiple files with regular-expression

2 Answers2