Bash script to remove 'x' amount of characters the end of multiple filenames in a directory?

Question

I have a list of file names in a directory (/path/to/local). I would like to remove a certain number of characters from all of those filenames.

Answer 1

This sed command will work for all the examples you gave.

sed -e 's/\(.*\)_.*\.moc1/\1_.moc1/'

However, if you just want to specifically "remove 5 characters before the last extension in a filename" this command is what you want:

sed -e 's/\(.*\)[0-9a-zA-Z]\{5\}\.\([^.]*\)/\1.\2/'

You can implement this in your script like so:

for filename in /path/to/local/*.moc1; do

    mv $filename "$(echo $filename | sed -e 's/\(.*\)[0-9a-zA-Z]\{5\}\.\([^.]*\)/\1.\2/')";

done

First Command Explanation

The first sed command works by grabbing all characters until the first underscore: \(.*\)_

Then it discards all characters until it finds .moc1: .*\.moc1

Then it replaces the text that it found with everything it grabbed at first inside the parenthesis: /\1

And finally adds the .moc1 extension back on the end and ends the regex: .moc1/

Second Command Explanation

The second sed command works by grabbing all characters at first: \(.*\)

And then it is forced to stop grabbing characters so it can discard five characters, or more specifically, five characters that lie in the ranges 0-9, a-z, and A-Z: [0-9a-zA-Z]\{5\}

Then comes the dot '.' character to mark the last extension : \.

And then it looks for all non-dot characters. This ensures that we are grabbing the last extension: \([^.]*\)

Finally, it replaces all that text with the first and second capture groups, separated by the . character, and ends the regex: /\1.\2/

Answer 2

 mv $filname $(echo $filename | sed -e 's/.....\.moc1$//');

or

 echo ${filename%%?????.moc1}.moc1

%% is a bash internal operator...

Answer 3

This might work for you (GNU sed):

sed -r 's/(.*).{5}\./\1./' file