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PHP Build a Simple PHP Application Wrapping Up The Project Objects

bruinebeer
bruinebeer
16,101 Points
Posted by bruinebeer
bruinebeer
16,101 Points

I cant get past this point. Changing $checker->number to 16661, just does not work. Adding it to code doenst eithe

I cant get past this point.

Changing $checker->number to 16661, just does not work.

Adding it to code doenst work either, what is wrong, i cant find it.

palprimes.php 
<?php

include("class.palprimechecker.php");
$checker = new PalprimeChecker;

$checker->number = 16661;


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

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


?>

2 Answers

Christian Andersson
Christian Andersson
8,712 Points

Your problem is with the constructor just above it. It should be:

$checker = new PalprimeChecker();

bruinebeer
bruinebeer
16,101 Points
bruinebeer
bruinebeer
16,101 Points

thanks that helped me out alot