[snip] I have a variable containing a string that contains the names of variables, and want to output the variable with the variables it contains evaluated. E.g.
$foo contains 'cat' $bar contains 'Hello $foo' and I want to output $bar as Hello cat The problem is that if I use echo $bar I just get Hello $foo [/snip] $bar = 'Hello' . $foo; -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
