Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

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



$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


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!