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PHP Build a Simple PHP Application Wrapping Up The Project Objects

Diego Villaseñor
Diego Villaseñor
12,615 Points

Stuck on challenge task 5

It gives me an error saying than isPalprime is undefined, I don't know what to do.

Thanks in advance

palprimes.php
<?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
Erik McClintock
45,783 Points

Diego,

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