Shell Programming and Scripting

Hi scripters,

I know sed can do this, but I'm not to well trained to achieve this.

Problem: wherever I am in my filesystem, with a 'pwd' I can check where I am.
It may be "/usr/local/apache/etc/conf" or "/opt/apache/var/log/backup" or "/var/log/backup/apache/2009/december"

Now from these paths I want to catch the part up to "apache" and cut everyting that comes behind it, so I 'll get /usr/local/apache, /opt/apache and /var/log/backup/apache.
Can this be done?
If possible with a bit of explanation, please.

Thanks alot, Qube

Code:

pwd | sed 's/apache.*/apache/'

maybe with perl like this:

pwd|perl -ne'm/(.+apache)\/{0,1}/;print "$1\n"'

of course 'pwd' can replaced with anything that outputs the directory name or a list of directory names.

Yes.
Assuming your current directory is /usr/local/apache/etc/conf, the sed command searches for the string apache and all the stuff behind, finds apache/etc/conf in this case and then replaces it with the string apache.
HTH

A pure sh approach:

Code:

d=$PWD
case $d in
    */apache/*) d=${d%%/apache/*}/apache;;
esac

Regards,
Alister

Thank YOU!
Afterwards things always look surprisingly easy.

Just for the sake of theory I have an additional question.
What if the path has 2 occurences of "apache" and I want to chop off from the first occurence (like in my first question)?

Thanks, Qube

IMHO, the sed command from vgersh99 does already do that. Just try it.