> How can I make this assignment in one statement. I know its going to be
something easy.
$path=$5;
$path=~s/ /\\ /g; <
won't
$path=$5=~s/ /\\ /g;
work? maybe have to put () around the $5=~, but I think it will work without that.
-
To unsubscribe from this list: send the line "unsubscribe linu
Paul Kraus wrote:
>
> How can I make this assignment in one statement.
> I know its going to be something easy.
> $path=$5;
> $path=~s/ /\\ /g;
I don't understand. Is this part of a bash script?
It looks like you a trying to assign the fifth parameter
(path=$5) to a variable named path (which
Phillip Morgan wrote:
> Further to the dilemma regarding not being able to use perl scripts with
> the crypt() function in perl 5.6.1, the problem seems to be that despite
> the fact that libcrypt does exist (/lib/libcrypt-2.0.7.so, or the
> symbolic link /lib/libcrypt.so.1 which points to the 2
Phillip Morgan wrote:
>
> Hi,
>
> Actually, just had a thought,
>
> /lib was not in /etc/ld.so.conf. Could this have something to do with the
> problem of not finding /lib/libcrypt?
>
> I added the entry to ld.so.conf, but it made no difference.
>
> I suppose there is a command I have to run
Hi,
Actually, just had a thought,
/lib was not in /etc/ld.so.conf. Could this have something to do with the
problem of not finding /lib/libcrypt?
I added the entry to ld.so.conf, but it made no difference.
I suppose there is a command I have to run to tell the system I've updated
ld.so.conf?
