How to parse full path of file name in bash ?

hq4ever · 03-28-2005, 01:17 PM

Help please.

Assuming I have this file:

Code:

/tmp/dir001/somefile.foo.gz

and I need to rename it to it's md5sum, if it didn't had an extersion I would have done it like this:

Code:

for i in *; mv "$i" `md5sum "i$" | cut -d ' ' -f 1`; done

But I need to keep the last extersion, i.e. I need it to be names [md5sum-goes-here].foo.gz

How can this be done ?

Thank you for helping,
Maxim.

TheLinuxDuck · 03-28-2005, 02:25 PM

Here's an example that shows nothing more than how to determine filename without it's extension, and the extension. Assuming the script is called "filename_split.sh", and we're to pass in a single filename:

Code:

#!/bin/bash

if [ -z "$1" ]; then
  echo "$0 filename_to_tweak"
  exit
fi

#  lines of note
#
pathfilename=${1%.*}
fileext=${1##*.}
#
#

echo "Full: $1"
echo "Pre : $pathfilename"
echo "Post: $fileext"

And run:

Code:

~/bash> ./filename_split.sh gcrt1.o
Full: gcrt1.o
Pre : gcrt1
Post: o

Here's a link for some other interesting bash matching/splitting

hq4ever · 03-28-2005, 03:31 PM

Aha "Parameter Expansion".

Exactly what I've been looking for.
Thank you very much.