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..s         


        

Answer 1

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.

Answer 2

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 3

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 4

You forgot a dot in front of the asterisk:

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

On OpenSUSE, RedHat, Gentoo you have to use Perl version of rename. This answer shows how to obtain it. On Arch, the package is called perl-rename.

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

Really cool lil diddy. 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:

OldName.ext    NewName.ext
OldName.ext    NewName.ext
OldName.ext    NewName.ext
OldName.ext    NewName.ext

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

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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