$varname = \$_SERVER['REMOTE_ADDR'];
$varvalue = $$varname;
That's wrong. Offhand you'll end up printing a string. I tried this:
?php
$a = 365;
$b = 366;
$var = $_GET['var'];
echo $$var;
?
And it was fine.
--
Richard Heyes
HTML5 Graphing for FF, Chrome, Opera and
That's fine as a test, but you never want to get a variable name from a
URL in practice.
Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com
Richard Heyes wrote:
$varname = \$_SERVER['REMOTE_ADDR'];
$varvalue = $$varname;
That's wrong. Offhand you'll
That's fine as a test, but you never want to get a variable name from a
URL in practice.
Of course you can, as long as it's sanitized and checked.
--
Richard Heyes
HTML5 Graphing for FF, Chrome, Opera and Safari:
http://www.rgraph.org
--
PHP General Mailing List (http://www.php.net/)
To
I mean that it is open for hacking if you pass a variable name through a
URL.
Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com
daniel danon wrote:
What do you mean?
On Sun, Oct 12, 2008 at 5:40 PM, Micah Gersten [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
Hi, I was wondering,
By php.net manual, Please note that variable variables cannot be used with
PHP's Superglobal arrays within functions or class methods. Is there any
way to override this problem? Just the not nice eval(return $variable);?
and in simple words - is there any way to make the
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