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Top Forums UNIX for Dummies Questions & Answers passing a variable inside a variable to a function Post 302158253 by KingVikram on Monday 14th of January 2008 07:28:35 PM
Old 01-14-2008
Quote:
Originally Posted by gus2000
Ah, you want the shell to parse "$C$x" as "$C1" but instead it is doing "$C" + "$x". So, you need to force it to evaluate $x first.

Code:
# C1=abc
# x=1 
# echo $C$x
1
# eval echo \$C$x
abc

Is that what you were looking for?
Yes, I believe thats the way to do it, but how to call the function using the above ? (BTW, thanks for your post)
 

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SESSION_UNREGISTER(3)							 1						     SESSION_UNREGISTER(3)

session_unregister - Unregister a global variable from the current session

SYNOPSIS
bool session_unregister (string $name) DESCRIPTION
session_unregister(3) unregisters the global variable named $name from the current session. Warning This function has been DEPRECATED as of PHP 5.3.0 and REMOVED as of PHP 5.4.0. PARAMETERS
o $name - The variable name. RETURN VALUES
Returns TRUE on success or FALSE on failure. NOTES
Note If $_SESSION (or $HTTP_SESSION_VARS for PHP 4.0.6 or less) is used, use unset(3) to unregister a session variable. Do not unset(3)$_SESSION itself as this will disable the special function of the $_SESSION superglobal. Caution This function does not unset the corresponding global variable for $name, it only prevents the variable from being saved as part of the session. You must call unset(3) to remove the corresponding global variable. Caution If you are using $_SESSION (or $HTTP_SESSION_VARS), do not use session_register(3), session_is_registered(3) and session_unregis- ter(3). PHP Documentation Group SESSION_UNREGISTER(3)
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