Output a variable using a variable.

snorcha

Is it possible to reference a variable using another variable in the name.

for example..

just say i have five variables..

$d_1='dude';
$d_2='wotever';
$d_3='i rock, but im not made of stone';
$d_4='ffs work';
$d_5='carn ya carnie';

just say i want to use a for loop to output the vars..
$top=5;
for($i=1;$i < $top;$i++)
{
echo $d_$i;
}

it errors....... I have tried

echo $d_{$i};

but no go..

I cannot use an array for other reasons..

Is there a way of doing it?

Cheers..
Snorcha

jblack

Try doing it like this:

$d[0] = "dude";
$d[1] = "wotever";
$d[2] = "i rock, but i'm not made of stone";

$i = 0;
while($d[$i])
	{
	echo $d[$i];
	$i++;
	}

Weedpacket

${"d_$i"}, but why can't you use an array?

snorcha

I can't use an array as it is a form with an array within an array.. I have another problem.. here is an example of what I am trying to do..

if($check=='yes') 
{
	for($i=1;$i < 3;$i++)
		{ 
		$t_c=count(${"d_$i"}); 
		for ($t_i=0;$t_i<$t_c;$t_i++)
			{
			echo ${"d_$i[$t_i]"};
			echo '<br>';
			}
 }
 }

I can't seem to get it to output the required var..

Cheers in advance..

Sn0rcha

Weedpacket

${"d_$i"}[$t_i]

..., but you can't name the fields d[1][], d[2][].... etc?