How can I script to shorten a filename?

rsmccain · 04-09-2008, 04:26 PM

I have a file called foo123.txt.gpg. I need it to become foo123.txt.

I know how to append characters to a filename but how can I get rid of a few characters at the end of a filename?

Thx..

forrestt · 04-09-2008, 04:29 PM

What is it that is determining what you want to remove? The last 4 characters? The last . and everything after it? The first two strings separated by dots are kept and everything else after is removed?

Let us know,

Forrest

colucix · 04-09-2008, 04:34 PM

Code:

file=foo123.txt.gpg
mv $file ${file%.gpg}

rsmccain · 04-09-2008, 05:19 PM

Quote:

Originally Posted by forrestt

What is it that is determining what you want to remove? The last 4 characters? The last . and everything after it? The first two strings separated by dots are kept and everything else after is removed?

Let us know,

Forrest

There will always be a file thats starts with foo, followed by 3 numbers, then .txt.gpg. (ie: foo123.txt.gpg, foo245.txt.gpg, foo452.txt.gpg).

I need a script to see if foo* exists, if so, remove the .gpg from the end.

rsmccain · 04-09-2008, 05:20 PM

Quote:

Originally Posted by colucix

Code:

file=foo123.txt.gpg
mv $file ${file%.gpg}

What is the purpose of the %? Does that mean get rid of everything after %?

gilead · 04-09-2008, 05:54 PM

I'm assuming your shell is bash. If so, have a look at the man page for bash and search for "Parameter Expansion":

Code:

${parameter%word}
${parameter%%word}
              The word is expanded to produce a pattern just as in pathname expansion.  If the pat-
              tern  matches  a trailing portion of the expanded value of parameter, then the result
              of the expansion is the expanded value of parameter with the shortest  matching  pat-
              tern  (the ``%'' case) or the longest matching pattern (the ``%%'' case) deleted.  If
              parameter is @ or *, the pattern removal operation  is  applied  to  each  positional
              parameter in turn, and the expansion is the resultant list.  If parameter is an array
              variable subscripted with @ or *, the pattern removal operation is  applied  to  each
              member of the array in turn, and the expansion is the resultant list.

ghostdog74 · 04-09-2008, 08:54 PM

Code:

awk 'BEGIN{q="\047"}
{
 old=FILENAME;
 sub(".gpg","",FILENAME)
 cmd="mv "q old q" "q FILENAME q
 system(cmd)
}' foo*.gpg

rsmccain · 04-10-2008, 09:33 AM

Quote:

Originally Posted by gilead

I'm assuming your shell is bash. If so, have a look at the man page for bash and search for "Parameter Expansion":

Code:

${parameter%word}
${parameter%%word}
              The word is expanded to produce a pattern just as in pathname expansion.  If the pat-
              tern  matches  a trailing portion of the expanded value of parameter, then the result
              of the expansion is the expanded value of parameter with the shortest  matching  pat-
              tern  (the ``%'' case) or the longest matching pattern (the ``%%'' case) deleted.  If
              parameter is @ or *, the pattern removal operation  is  applied  to  each  positional
              parameter in turn, and the expansion is the resultant list.  If parameter is an array
              variable subscripted with @ or *, the pattern removal operation is  applied  to  each
              member of the array in turn, and the expansion is the resultant list.

Thanks for everyones responses. We ended up doing this....

Code:

for i in `find /tmp/ -name foo*pgp -print0`; do
newvar=`echo $i | sed s/.pgp//g`
mv $i $newvar
done

Now if someone googles with the same question, they have multiple options.

Thanks again.