Topic: Variable Variables

[sstangle73]

i am trying to make a script that will make a varible 18 times here is what i have something is wrong

for($i=1; $i<=18; $i++){
${"player" . $i . "_number"} = ${$array[player .  $i . _number]};
}

it is something with the second half the

Code: [Select]

${$array[player .  $i . _number]};i am trying to get it so it will do:

Code: [Select]

$player1_number = $array['player1_number'];
$player2_number = $array['player2_number'];
$player3_number = $array['player3_number'];
ect

Thanks!

EDIT
the error i am getting is

Code: [Select]

Notice: Use of undefined constant player - assumed 'player' in /homepages/27/d193007783/htdocs/maddogmania/application/roster.php on line 22

Notice: Use of undefined constant _number - assumed '_number' in /homepages/27/d193007783/htdocs/maddogmania/application/roster.php on line 22

Notice: Undefined variable: 07 in /homepages/27/d193007783/htdocs/maddogmania/application/roster.php on line 22
the third one where it has the variable 07, 07 is what the $player1_number should be set to if it worked right

[thorpe]

Assuming $array is an array.

${"player" . $i . "_number"} = $array['player' .  $i . '_number'];

You would be much better off using arrays instead of variable variables though. They are slow, and make your code much harder to read.

[sstangle73]

Thanks thorpe!!

[Andy-H]

if you already have the array keys set as your desired variable names, as in thorpe's example; look into extract().