I'm trying to use variable variables to work on arrays:
$forest = array("a", "b", "c", ...);
$layer[$l]= "forest";
Now I want to access all array members of $forest using $$layer:
e.g.
for($c = 0; $c < $$layer[$l]; $l++) {
echo $$layer[$l][$c];
}
But this doesn't work, gives syntax error,
So my workaround is:
$new = $$layer[$l];
$new[$c] refers all elements of array $forest
Is this the best workaround or am I missing something?
> I think you're missing the point of variable variables.
>
> <?
> $a = 'foo';
> $$a = 'bar';
>
> echo "$a $$a";
> ?>
>
> After the first use of $$a, you now have a variable called $foo with a
> value of 'bar'.
>
> So your echo would be echo "$a $foo";
>
> I kind of consider variable variables the poor mans array. Most any
> solution you think of with variable variables could be better solved by
> using arrays.
>
> ---John Holmes...
>
> > -----Original Message-----
> > From: Peter [mailto:[EMAIL PROTECTED]]
> > Sent: Wednesday, June 12, 2002 10:42 PM
> > To: Php
> > Subject: [PHP] Varible Varibles
> >
> > howdy,
> > I'm just curious here about varible varibles ... I know that you can,
> well
> > it's documented that you can, do the following
> >
> > <?
> > $a = foo;
> > $$a = bar;
> >
> > echo "$a $$a";
> > ?>
> >
> > which will produce foo bar
> >
> > now what I am curious about is, how much of a difference does that
> really
> > make when you compare it to..
> >
> > <?
> > $a = foo;
> > $a .= bar;
> >
> > echo "$a";
> > ?>
> >
> >
> >
> >
> > Cheers
> >
> > Peter
> > "the only dumb question is the one that wasn't asked"
> >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
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>
-Pushkar S. Pradhan
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