making vars in bash script accessible from shell

[Jerry Feldman]

This is a standard Unix feature. The parent process never inherits anything 
from a child. 
Shell variables are local. 
That said, the only way for a process to inherit variables from a script is 
to source the script. 
Bash supports the 'source' built-in command as well as the . builtin.
On 7 Nov 2002 at 15:19, Price, Erik wrote:

> Hi,
> 
> I know there's some hardcore shell heads on this list, so I was hoping I could solicit some assistance here... 
> 
> I've isolated the problem to an inability to make a variable declaration accessible outside of the script.  It's almost as if there were a bash form of "use strict" going on that was keeping my variables within the scope of the script.  If I try to echo $newPWD, I get nothing -- but I can see 
that $newPWD is getting set in the script.  I've tried "export newPWD", which doesn't do anything either.
> 
> This is obviously some fundamental rule of variable scope in bash that I just don't know, so can someone set me right?
> 
> Bash v. 2.02 on Solaris.
> 
> Thanks,
> Erik
> 
> 
> 
> #!/usr/local/bin/bash
> #(yes, that's the location of bash on my system)
> 
> # How many characters of the $PWD should be kept
> pwdmaxlen=30
> 
> # indicator that there has been directory truncation
> trunc_symbol="..."
> if [ ${#PWD} -gt $pwdmaxlen ]
> then
>         pwdoffset=$(( ${#PWD} - $pwdmaxlen ))
>         newPWD="${trunc_symbol}${PWD:${pwdoffset}:${pwdmaxlen}}"
> else
>         newPWD=${PWD}
> fi
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-- 
Jerry Feldman <gaf at blu.org>
Associate Director
Boston Linux and Unix user group
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