Variable variable. Read the docs.
$v = 'foo';
$foo = 'bar';
echo $$v;
Regards,
Andrey
P.S.
Sometimes {} are used : ${$v}
Scott Fletcher [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
The script was working great before PHP 4.2.x and not after that. So, I
looked through the
Interesting! Look like the 2nd $ is decomissioned and is reserve for
something in the future or something. Just like the _ is when it come
with $_POST as an example. That would explain why it doesn't work with PHP
4.2.x up.
Andrey Hristov [EMAIL PROTECTED] wrote in message
The cause for your problem would be that register_globals defaults to off in
PHP 4.2.x and greater.
The solution? Start using the new superglobals ($_POST, $_GET, $_SESSION
etc) or (not recomended) set register_globals = on in php.ini
Read more here:
variable variable... right up there with array array
basically what you are saying is resolve $var, then find out what that variable
holds
example;
assume your $counter is currently at 5
$var = v.$counter._high_indiv;
would mean that $var= v5_high_indiv
assuming that v5_high_indiv is
I tried that test script you mentioned and it doesn't work in PHP 4.2.1. I
have a very good idea why is that, must have to do with the php.ini.
Unfortunately, it doesn't work either. I'll tell you what, I'll just throw
out that script and write a different script. This time, no double $.
What docs at php.net? under variable, predefine variable or what?
Andrey Hristov [EMAIL PROTECTED] wrote in message
002601c22cd0$b1995170$1601a8c0@nik">news:002601c22cd0$b1995170$1601a8c0@nik...
Variable variable. Read the docs.
$v = 'foo';
$foo = 'bar';
echo $$v;
Regards,
Andrey
P.S.
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 like
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