note 9699 modified in Variable variables by jeroen

[Editors note: This doesn't work, because you try to assign a value to the variable named 'a[0]'. If you want this kind of behaviour (execute PHP-syntax out of strings), take a look at eval() ] You can't create an array using variable variables! example: $n="a" $$n=12 echo($a); this will work, hence, $a = 12; however, $n="a[0]"; $$n=12; $n="a[1]"; $$n=24; echo(count($a).",".$a[0].",".$a[1]); I expected the output to be: 2,12,24 However, actually this output nothing but three ","s. http://www.php.net/manual/variable-variables.php