why then the code doesn't work? the error I'm getting is: "Fatal error: Call to undefined function solution1() in ..." even the function itself is just above the line that calls function?
-afan Stut wrote: > Jay Blanchard wrote: >> I don't think you can put a function name in a variable and call it like >> $function($var). > > Yes you can - it's basically the same as variable variables. > > -Stut > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
