Regex help

Penguin of Wonder · 08-05-2007, 09:20 PM

I'm trying to split up a URL that matches a regex expression but I can't figure it out.

For example, if this were my URL:

Quote:

http://slackware.osuosl.org/slackwar...4.0-i486-1.tgz

How would I find and then chop off just the file name? With the above example all I would want left would be

Quote:

subversion-1.4.0-i486-1.tgz

So far the best I can do is

Code:

cat packagelist | 
while read line; do 
echo ${line} | grep '.*[.].*$'
done

I know I'm missing awk or sed or something. Thanks in advance!

Tinkster · 08-05-2007, 09:28 PM

Code:

sed -r 's@.+/([^/]+)$@\1@'

Cheers,
Tink

ta0kira · 08-05-2007, 09:44 PM

This is what the basename call/function are designed for.

Code:

basename http://slackware.osuosl.org/slackware-11.0/slackware/d/subversion-1.4.0-i486-1.tgz

ta0kira

Penguin of Wonder · 08-05-2007, 10:47 PM

Thanks for the quick response guys! Helped out a lot!

makyo · 08-06-2007, 07:04 AM

Hi.

Also built into bash:

Quote:

${parameter#word}
${parameter##word}
The word is expanded to produce a pattern just as in pathname
expansion. If the pattern matches the beginning of the value of
parameter, then the result of the expansion is the expanded
value of parameter with the shortest matching pattern (the ``#''
case) or the longest matching pattern (the ``##'' case) deleted.

-- excerpt from man bash

As in this example:

Code:

#!/bin/sh

# @(#) s1       Demonstrate parameter expansion in bash.

set -o nounset
echo
echo "GNU bash $BASH_VERSION" >&2
echo

P=${1-"http://slackware.osuosl.org/slackware-11.0/slackware/d/subversion-1.4.0-i486-1.tgz"}

base=${P##*/}

echo " basename: $base"

exit 0

Producing:

Code:

% ./s1

GNU bash 2.05b.0(1)-release

 basename: subversion-1.4.0-i486-1.tgz

cheers, makyo