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PHP

cathymacars
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cathymacars
Full Stack JavaScript Techdegree Student 15,941 Points
Posted by cathymacars
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cathymacars
Full Stack JavaScript Techdegree Student 15,941 Points

Build a Simple PHP Application > Wrapping Up The Project > Objects: Task 5/7

Hello, I'm having trouble with the Code Challenge mentioned in the title.

The question is: "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”."

My answer was:

<?php
include("class.palprimechecker.php");
$checker = new PalprimeChecker;
$checker->number = 17;
$number = $checker->number;
echo "The number " .$number. " ";
if(isPalprime($number)){
    echo "is";
} else {
    echo "is not";
}
echo " a palprime.";
?>

The error I'm getting: "PHP Fatal error: Call to undefined function isPalprime() in palprimes.php on line 15 The number 17 "

Help?

9 Answers

Randy Hoyt
STAFF
Randy Hoyt
Treehouse Guest Teacher
Randy Hoyt
Randy Hoyt
Treehouse Guest Teacher

So close! The problem is in this line:

if(isPalprime($number)){

Two things:

  1. You are calling isPalprime as if it is general function. It's not: it's a method of all PalprimeChecker objects. (Hint: You'll need to use the two characters that together look like a single arrow.)
  2. The question states that the isPalprime method does not receive any arguments, but you are passing in one argument. (It will know to check if 17 is a palprime when you call the isPalprime method because you set it as a property of the object.)

Does that help?

Steven Walters
Steven Walters
10,573 Points
Steven Walters
Steven Walters
10,573 Points

This quiz makes me appreciate JavaScript.

Michael Dion
Michael Dion
6,232 Points
Michael Dion
Michael Dion
6,232 Points

Glad I'm not the only one who struggled here. These challenges are definitely forcing me to have to (somewhat) understand what's going on instead of just copying what's in the video :-).

cathymacars
seal-mask
.a{fill-rule:evenodd;}techdegree
cathymacars
Full Stack JavaScript Techdegree Student 15,941 Points
cathymacars
.a{fill-rule:evenodd;}techdegree
cathymacars
Full Stack JavaScript Techdegree Student 15,941 Points

So, yeah, I noticed that it didn't receive any arguments. But since I tried all kinds of combinations like

if(isPalprime->$number))

or

if(isPalprime()->$checker ))

And nothing worked... I tried using it like a function, just to be sure... But it wasn't until you said "because you set it as a property of the object" that I understood! So I tried:

if($checker->isPalprime())

And it worked. Thanks! :D

Bryan Knight
Bryan Knight
34,215 Points
Bryan Knight
Bryan Knight
34,215 Points

I was SO stuck here as well. I couldn't figure it out to save my life. I read through this post three times and then it clicked. I'll definitely have to spend more time working with objects, methods and properties to make sure I have this nailed down.

Mauro Bonucci
Mauro Bonucci
5,953 Points
Mauro Bonucci
Mauro Bonucci
5,953 Points

I was stucked to in this one, but what was wrong in mine was the conditional, instead puttin "==" i was puttin "="

```echo "The number" . $checker->number; if(!$checker -> isPalprime() == $checker->number){

echo "is not";} echo " a palprime.";```

KJ Prince
KJ Prince
5,145 Points

It's unnecessary to use the ! operator in this question. You're checking if it is not equal to instead of just checking whether the number is palinprime or not using he conditional if statement to do the work.

here's what worked for me:

<?php

include ("class.palprimechecker.php"); $checker = new PalprimeChecker; $checker->number = 17;

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

echo " a palprime.";

?>

Martina Carrington
Martina Carrington
15,754 Points
Martina Carrington
Martina Carrington
15,754 Points

i am really having problem with Challenge Task 7 of 7

Jason S
Jason S
16,247 Points
Jason S
Jason S
16,247 Points

just put 151 where 16661 is

Stephen Garner
PLUS
Stephen Garner
Courses Plus Student 11,241 Points
Stephen Garner
Stephen Garner
Courses Plus Student 11,241 Points

If anyone is having trouble...my advice is to complete the first time, and then do it a second time. New concepts that you have a difficult time understanding must be practiced...

I have re-completed quizzes and assignments because I had difficulty completing them. Some people learn there repetition.

$checker->isPalprime()

Micah Doulos
Micah Doulos
17,663 Points
Micah Doulos
Micah Doulos
17,663 Points

The important thing to remember is that isPalprime() is not a global function. If you think of it is as a function, it would be a function contained within or inside the object called $checker. So it does not make sense to try to call isPalprime($checker). The isPalprime method is itself subject to the $checker object and thus must be called inside of the object, not vice-versa.