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PHP Build a Simple PHP Application Wrapping Up The Project Objects

Posted by Raul Chiru
Raul Chiru
Courses Plus Student 2,776 Points

hard thing.. help, please

PalprimeChecker objects have a method called isPalprime(). This method does not receive any arguments. It returns true if the number property contains a palprime, and it returns false if the number property does not contain a palprime. (Tip: 17 is not a palprime.) Turn the second echo statement into a conditional that displays “is” or “is not” appropriately. The conditional should call the isPalprime method of the $checker object. If the isPalprime method returns true, then echo “is”; otherwise, echo “is not”.

palprimes.php 
<?php
include('class.palprimechecker.php');    
$checker = new PalprimeChecker(){
 function isPalprime(); 
}
$checker->number = 17;                     




echo "The number ".$checker->number;    
if($checker == isPalrime()){
 echo "is";
} else
  echo "is not";
echo " a palprime.";

?>

1 Answer

Grace Kelly
Grace Kelly
33,990 Points
Grace Kelly
Grace Kelly
33,990 Points

Hi Raul, you are definitely on the right track!! However you do not need to declare the isPalprime() method as it already belongs to the PalprimeChecker object, and you have included it with the include("class.palprimechecker.php") function!! All you need to do is remove it from your $checker declaration and it should work :)

<?php

include("class.palprimechecker.php");
$checker = new PalprimeChecker();
$checker->number = 17;


echo "The number " .$checker->number;

if ($checker->isPalprime()) {
    echo "is";
}
else {
    echo "is not";
}
echo " a palprime.";

?>

Hope that helps!!

Raul Chiru
Raul Chiru
Courses Plus Student 2,776 Points

Thank you so much, Grace!

Grace Kelly
Grace Kelly
33,990 Points
Grace Kelly
Grace Kelly
33,990 Points

No problem Raul glad to help! happy coding!! :)