[plug] Shell Scripting Confusion

[Chris Griffin]

Because when $output is empty (null) it evaluates to: "if [ -n ]; then " as 
Mike showed.
But when you put the quotes around it, it correctly evaluates to:

"if [ -n ""]; then "

Regards,
Chris


At 05:26 9/02/2000 +0000, you wrote:
>Mike Holland wrote:
> >
> > On Wed, 9 Feb 2000, Christian wrote:
>
> > When you see the answer you'll want to stick with perl.
> > Shell programming is ugly.
>
>Well, I do really want to stick with Perl but I'm hacking one of the
>scripts on the system so while it's very tempting to just do "perl -e
>'<insert nice, sensible, sane code in here>'", it doesn't look that good
>really, does it? :-)
>
> > > if [ -n $output ]; then
> >
> > Since $output is empty, that evaluates as  "if [ -n ]; then "
> > which ought to at least give an error message?
> >
> > Try:
> >         if [ -n "$output" ]; then
> >
> > I'll bet you're groaning by now.
>
>Hmmm... yeah, a little... but why does $output evaluate to being empty
>whereas "$output" has meaning?  Is it that you require the quotes to
>actually turn it into a string?
>
>Thanks for your help,
>
>Christian.

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