Re: [PHP] 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" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > Interesting! Look l

Re: [PHP] Purpose of $$var ?????

What docs at php.net? under variable, predefine variable or what? "Andrey Hristov" <[EMAIL PROTECTED]> wrote in message 002601c22cd0$b1995170$1601a8c0@nik">news:002601c22cd0$b1995170$1601a8c0@nik... > Variable variable. Read the docs. > > $v = 'foo'; > $foo = 'bar'; > echo $$v; > > Regards, > An

Re: [PHP] Purpose of $$var ?????

I tried that test script you mentioned and it doesn't work in PHP 4.2.1. I have a very good idea why is that, must have to do with the php.ini. Unfortunately, it doesn't work either. I'll tell you what, I'll just throw out that script and write a different script. This time, no double "$". --c

RE: [PHP] Purpose of $$var ?????

variable variable... right up there with array array basically what you are saying is resolve $var, then find out what that variable holds example; assume your $counter is currently at 5 $var = "v".$counter."_high_indiv"; would mean that $var= "v5_high_indiv" assuming that v5_high_ind

RE: [PHP] Purpose of $$var ?????

The cause for your problem would be that register_globals defaults to off in PHP 4.2.x and greater. The solution? Start using the new superglobals ($_POST, $_GET, $_SESSION etc) or (not recomended) set register_globals = on in php.ini Read more here: http://www.php.net/manual/en/language.variable

Re: [PHP] 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" <[EMAIL PROTECTED]> wrote in message 002601c22c

Re: [PHP] Purpose of $$var ?????

Variable variable. Read the docs. $v = 'foo'; $foo = 'bar'; echo $$v; Regards, Andrey P.S. Sometimes {} are used : ${$v} "Scott Fletcher" <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>... > The script was working great before PHP 4.2.x and not after that. So, I > looked thro