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PHP Build a Simple PHP Application Wrapping Up The Project Objects

Posted by Rich Mathis
Rich Mathis
3,626 Points

Hung up on objects question 5 of 7

Here is the challenge...

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”.

This is what I have and I can't figure out what the issue is.

<?php

include ('class.palprimechecker.php');

$checker = new PalprimeChecker; 
$checker->number = 17;

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

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

?>

2 Answers

Bella Ratmelia
.a{fill-rule:evenodd;}techdegree seal-36
Bella Ratmelia
Front End Web Development Techdegree Graduate 28,947 Points

The question said "PalprimeChecker objects have a method called isPalprime()" $checker is a PalprimeChecker object. Therefore, to call the method that belongs to that object, you can simply do it like this:

$checker->isPalprime()

Therefore your code should be:

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

Hope this helps!

Rich Mathis
Rich Mathis
3,626 Points
Rich Mathis
Rich Mathis
3,626 Points

Thanks! I was staring at that for a long time and it finally makes sense.