I have this code
<?php
$a1 = 'http://www.hamooz.com';
$a2 = 'http://www.myegy.com';
$a3 = 'http://www.tech-wd.com/wd';
$num = rand(1,3);
$numb = '$a'.$num;
echo file_get_contents($numb);
?>
the problem appear like:
Warning: file_get_contents($a3) [function.file-get-contents]: failed to open stream: No such file or directory in C:\xampp\htdocs\ttt\bessah.php on line 9
Variable variables are a bad practice. Use an array instead.
<?php
$a = array();
$a[] = 'http://www.hamooz.com';
$a[] = 'http://www.myegy.com';
$a[] = 'http://www.tech-wd.com/wd';
$num = rand(0,2);
$numb = $a[$num];
echo file_get_contents($numb);
?>
For this specific task however, there is also a shortcut:
array_rand ( )
Picks one or more random entries out of an array, and returns the key (or keys) of the random entries.
You need to do $numb = ${'a'.$num};, but use the array solution as @Pekka's answer.
I agree with @Pekka you're going to complicate your code, but if you absolutly want to use dynamic variables you can do :
<?php
$a1="hello";
$b=1;
echo ${"a$b"};
?>