Welcome to the Treehouse Community

The Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.

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!