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 <[EMAIL PROTECTED]>
Associate Director
Boston Linux and Unix user group
http://www.blu.org PGP key id:C5061EA9
PGP Key fingerprint:053C 73EC 3AC1 5C44 3E14 9245 FB00 3ED5 C506 1EA9
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