Rename files using regular expression in linux

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)

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.

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