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

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 > > > > >