Hi Jeff.
$varname = C_First_Name.$x;
// e.g. if $x == 3 then ...
echo $$varname; // ... prints the content of $C_First_Name3
Hope this is what you wanted ?!
Greetinx,
Mike
Michael Rudel
- Web-Development, Systemadministration -
Besuchen Sie uns am 20. und 21. August 2001 auf der
Hi, I'm not quite getting it. If $x is 1 here is what happens.
$varname = C_First_Name.$x;
echo $$varname;
then $$varname prints as $C_First_Name1. However $C_First_Name1
prints the contents of the variable.
Jeff Oien
Hi Jeff.
$varname = C_First_Name.$x;
// e.g. if $x == 3 then ...
I found a solution but not sure if it's the most efficient.
I need to first define a simpler variable:
$C_First_Name = ${C_First_Name.$x};
then INSERT VALUE $C_First_Name.
Jeff Oien
Hi Jeff.
$varname = C_First_Name.$x;
// e.g. if $x == 3 then ...
echo $$varname; // ... prints the
