export multiple executable paths in a parent directory using regular expressions?

lmcilwain · 10-02-2012, 01:47 PM

Hello,

I am wondering if there is a way in bash to export an executable path using regular expressions. What I am attempting to do is organize all my executable code into folders but don't want to add each folder to the executable path in my .bash_profile when I create a new one.

Can anyone tell me if it is possible to add something like the following example to my .bash_profile:

Code:

PATH=/home/user/scripts/*;
export PATH

So that I could have execute a file that is possibly located in

Code:

/home/user/scripts/some_other_folder/some_script.pl

pan64 · 10-03-2012, 07:13 AM

A=(`ls -1s /home/user/scripts/*`)
echo "${A[*]/ /:}"
will print what you need, but will not be checked if there were any files

catkin · 10-03-2012, 09:01 AM

How about something like

Code:

export PATH=$( find -regex <your regex> -type d -printf '%p:' )

That ends the list with a trailing : which may not be what you want in which case try something like:

Code:

path=$( find -regex <your regex> -type d -printf '%p:' )
export PATH=${path%:}