Rename files using regular expression in linux

Question

I have a set of files named like:

Friends - 6x03 - Tow Ross' Denial.srt
Friends - 6x20 - Tow Mac and C.H.E.E.S.E..srt
Friends - 6x05 - Tow Joey's Porshe.srt

and I want to rename them like the following

S06E03.srt
S06E20.srt
S06E05.srt

what should I do to make the job done in linux terminal? I have installed rename but U get errors using the following:

rename -n 's/(\w+) - (\d{1})x(\d{2})*$/S0$2E$3\.srt/' *.srt

Answer 1

find + perl + xargs + mv

xargs -n2 makes it possible to print two arguments per line. When combined with Perl's print $_ (to print the $STDIN first), it makes for a powerful renaming tool.

find . -type f | perl -pe 'print $_; s/input/output/' | xargs -d "\n" -n2 mv

Results of perl -pe 'print $_; s/OldName/NewName/' | xargs -n2 end up being:

OldName1.ext    NewName1.ext
OldName2.ext    NewName2.ext
OldName3.ext    NewName3.ext
OldName4.ext    NewName4.ext

I did not have Perl's rename readily available on my system.

---

How does it work?

1. find . -type f outputs file paths (or file names...you control what gets processed by regex here!)
2. -p prints file paths that were processed by regex, -e executes inline script
3. print $_ prints the original file name first (independent of -p)
4. -d "\n" cuts the input by newline, instead of default space character
5. -n2 prints two elements per line
6. mv gets the input of the previous line

---

My preferred approach, albeit more advanced.

Let's say I want to rename all ".txt" files to be ".md" files:

find . -type f -printf '%P\0' | perl -0 -l0 -pe 'print $_; s/(.*)\.txt/$1\.md/' | xargs -0 -n 2 mv

The magic here is that each process in the pipeline supports the null byte (0x00) that is used as a delimiter as opposed to spaces or newlines. The first aforementioned method uses newlines as separators. Note that I tried to easily support find . without using subprocesses. Be careful here (you might want to check your output of find before you run in through a regular expression match, or worse, a destructive command like mv).

How it works (abridged to include only changes from above)

1. In find: -printf '%P\0' print only name of files without path followed by null byte. Adjust to your use case-whether matching filenames or entire paths.
2. In perl and xargs: -0 stdin delimiter is the null byte (rather than space)
3. In perl: -l0 stdout delimiter is the null byte (in octal 000)

Answer 2

Edit: found a better way to list the files without using IFS and ls while still being sh compliant.

I would do a shell script for that:

#!/bin/sh
for file in *.srt; do
  if [ -e "$file" ]; then
    newname=`echo "$file" | sed 's/^.*\([0-9]\+\)x\([0-9]\+\).*$/S0\1E\2.srt/'`
    mv "$file" "$newname"
  fi
done

---

Previous script:

#!/bin/sh
IFS='
'
for file in `ls -1 *.srt`; do
  newname=`echo "$file" | sed 's/^.*\([0-9]\+\)x\([0-9]\+\).*$/S0\1E\2.srt/'`
  mv "$file" "$newname"
done

Answer 3

Use mmv (mass-move?)

It's simple but useful: The * wildcard matches any string (without slashes) and ? matches any character in the string to be matched. Use #X in the replace string to refer to the X-th wildcard match.

In your case:

mmv 'Friends - 6x?? - Tow *.srt' 'S06E#1#2.srt'

Here #1#2 represent the two digits which are captured by ?? (match #1 and #2).
So the following replacement is made:

Friends - 6x?? - Tow *           .srt    matches
Friends - 6x03 - Tow Ross' Denial.srt    which is replaced by
            ??
        S06E03.srt

mmv also offers matching by [ and ] and ;.

You can not only mass rename, but also mass move, copy, append and link files.

See the man page for more!

Personally, I use it to pad numbers such that numbered files appear in the desired order when sorted lexicographically (e.g., 1 appears before 10): file_?.ext ? file_0#1.ext

Answer 4

Not every distro ships a rename utility that supports regexes as used in the examples above - RedHat, Gentoo and their derivatives amongst others.

Alternatives to try to use are perl-rename and mmv.

Answer 5

if your linux does not offer rename, you could also use the following:

find . -type f -name "Friends*" -execdir bash -c 'mv "$1" "${1/\w+\s*-\s*(\d)x(\d+).*$/S0\1E\2.srt}"' _ {} \;

i use this snippet quite often to perform substitutions with regex in my console.

i am not very good in shell-stuff, but as far as i understand this code, its explanation would be like: the search results of your find will be passed on to a bash-command (bash -c) where your search result will be inside of $1 as source file. the target that follows is the result of a substitution within a subshell, where the content of $1 (here: just 1 inside your parameter-substituion {1//find/replace}) will also be your search result. the {} passes it on to the content of -execdir

better explanations would be appreciated a lot :)

please note: i only copy-pasted your regex; please test it first with example files. depending on your system you might need to change \d and \w to character classes like [[:digit:]] or [[:alpha:]]. however, \1 should work for the groups.

Answer 6

I think the simplest as well as universal way will be using for loop sed and mv. First, you can check your regex substitutions in a pipe:

ls *.srt | sed -E 's/.* ([0-9])x([0-9]{2}) .*(\.srt)/S\1E\2\3/g'

If it prints the correct substitution, just put it in a for loop with mv

for i in $(ls *.srt); do 
    mv $i $(echo $i | sed -E 's/.* ([0-9])x([0-9]{2}) .*(\.srt)/S\1E\2\3/g') 
    done

Answer 7

You can use rnm:

rnm -rs '/\w+\s*-\s*(\d)x(\d+).*$/S0\1E\2.srt/' *.srt

Explanation:

1. -rs : replace string of the form /search_regex/replace_part/modifier
2. (\d) and (\d+) in (\d)x(\d+) are two captured groupes (\1 and \2 respectively).

More examples here.