Echo 1/3 of $var

[alirezaok]

hi
at this code i want if my $var has not * character its contenet it shows 1/3 of contenet of $var.

$var="salam chetori khoobi ya na";

if(strpos($var, "*") === false)
{
$len = settype(strlen($var)/3, "integer");
echo substr($var, 0, $len);
}

but it is not work and showes this error:

Fatal error: Only variables can be passed by referenc

anyone can reform it?

[bubblenut]

change
$len = settype(strlen($var)/3, "integer");
to
$len= (int)(strlen($var)/3);

HTH
Bubble

[Shrike]

Something like

<?php
$var = "test * test";
if(strpos($var, "*") === FALSE){
	echo substr($var, 0, round(strlen($var)/3));
}else{
	echo $var;
}
?>

Race you to the next one Bubble...lol