Re: Purpose of $$var ?????

Alright! Found the problem! Faulty script written that come before this script where $$var come into play. At least, it wasn't me, it was the other programmer's error. :-) "Scott Fletcher" <scott@abcoa.com> wrote in message news:20020716140950.61894.qmail@pb1.pair.com... > Interesting! Look like the 2nd "$" is decomissioned and is reserve for > something in the future or something. Just like the "_" is when it come > with $_POST as an example. That would explain why it doesn't work with PHP > 4.2.x & up. > > "Andrey Hristov" <ahristov@icygen.com> wrote in message > news:002601c22cd0$b1995170$1601a8c0@nik... > > Variable variable. Read the docs. > > > > $v = 'foo'; > > $foo = 'bar'; > > echo $$v; > > > > Regards, > > Andrey > > > > P.S. > > Sometimes {} are used : ${$v} > > > > > > > > > > "Scott Fletcher" <scott@abcoa.com> wrote in message > > news:<20020716135017.43862.qmail@pb1.pair.com>... > > > The script was working great before PHP 4.2.x and not after that. So, I > > > looked through the code and came upon this variable, "$$var". I have no > > > idea what the purpose of the double "$" is for a variable. Anyone know? > > > > > > --clip-- > > > $var = "v".$counter."_high_indiv"; > > > $val3 = $$var; > > > --clip > > > > > > Thanks, > > > FletchSOD > > > > > > > > > > > > -- > > > PHP General Mailing List (http://www.php.net/) > > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > > > > > >