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Start your free trialDiego Villaseñor
12,615 PointsStuck on challenge task 5
It gives me an error saying than isPalprime is undefined, I don't know what to do.
Thanks in advance
<?php
include("class.palprimechecker.php");
$checker = new PalprimeChecker;
$checker -> Number = 17;
$checker -> isPalprime();
echo "The number " . 17 . " ";
if (isPalprime($checker)) {
echo "is";
} else {echo "is not";}
echo " a palprime.";
?>
1 Answer
Erik McClintock
45,783 PointsDiego,
There are two problems here.
1) The isPalprime is a method on a PalprimeChecker object, so you need to call it as such using the syntax we've seen before for PHP objects: the "->" symbol
2) The instructions inform us that the isPalprime method does not accept any arguments, but you are trying to pass in the $checker object as an argument. Instead, call this method ON the $checker object
After making those changes, you should be moving right along!
Erik