Getting the first part of a filename in a BASH script

trevelluk · 02-14-2005, 09:58 AM

Hi all, hopefully someone will be able to help with a little shell scripting problem.

Basically, I need to be able to find out the "first part" of a filename, i.e. the bit before the extension. I'm only looking at this for .tar.gz files (so if, say, there was a file called myfile.tar.gz, I'd want to get the result myfile). The first solution that came to mind was

Code:

cut -d'.' -f1

but this doesn't work if there's a dot anywhere else in the name (like my.file.tar.gz). If anyone's got any ideas, they'll be much appreciated.

ilikejam · 02-14-2005, 10:08 AM

Hi.

Try:
basename myfile.tar.gz .tar.gz

Dave

trevelluk · 02-14-2005, 10:15 AM

Great, that works perfectly! Thanks.

jschiwal · 02-15-2005, 01:06 AM

If the filename is in a variable you could use variable substitution.

> bn=myfile.tar.gz
> echo ${bn%%.*}
myfile

This could be handy in scripts.
Now, if the filename contains a dot character, then this won't work. If you want to remove only the last extension, use ${variable-name%.*} instead.