Updating $_POST var, sprintf problem

Enlightened

Hello, Im updating a post var using sprintf to add 2.00 to it. However it will knock off the value after the decimal and replace it with .00

for instance

// $_POST['amount'] = 100.22

$_POST['amount'] = sprintf('%.2f', (int) $_POST['amount']+2);

echo $_POST['amount'];

// returns 102.00

I would like it to output "102.22"

Am i not using the proper syntax?

Any help is much appreciated.

devinemke

Enlightened wrote:
However it will knock off the value after the decimal and replace it with .00

because your are casting the variable as an integer by using the (int) syntax. if you simply want to add 2 then you do not need sprintf at all.

$amount = 100.22;
$amount += 2;
echo $amount;

Enlightened

Thanks for your reply.

I remember when i tried that, the $_POST would come back empty.

It did work with sprintf, Maybe i should try to remove

the (int), Or completely do away with sprintf?

mmonaco

Just add 2.00

Enlightened

ok i tried those suggestions, they all yield the same

// $_POST['amount'] = 100.99

$_POST['amount'] += 2;

// $_POST['amount'] now equals 102.00

This is very wierd, anyone else experience this?

Rodney_H_

What happens when you try this simple test:

<?php
if(isset($_POST['submit'])){
	echo 'original amount: ' . $_POST['amount'] . '<br />';
	// add 2 to post amount
	$_POST['amount'] += 2;
	// echo new value
	echo 'new amount: ' . $_POST['amount'];
}
else {
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" name="amount" size="25" /> Amount
<br />
<input type="submit" name="submit" value="submit" />
</form>
<?php
}
?>

Do you get the correct values??

halojoy

Enlightened wrote:
Maybe I should try to remove
the (int), Or completely do away with sprintf?

(int) POST['amount'] will be 100
when you add 2 you get 102

Remove (int)
which makes an integer out of posted value

I have not tested, but think you have your problem there
like you have also mentioned in your above quote