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php undef function

Swiebertjeee
By Swiebertjeee
in Programming
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Go to solution Solved by dragosudeki,

I don't know PHP, but aren't you supposed to do it like this?

object->function()

Its a function of an object, like getNaam(). The error is occuring because it assumes you made a printGegevens() function, but you haven't. The printGegevens() is supplied under an object, but you are not using the correct syntax to use that function.

Posted

I am having some trouble with PHP I can't find out why it wont work.. 

 

I have the following error: 

Fatal error: Call to undefined function printGegevens() in C:\xampp\htdocs\OOP\persoon.php on line 29

this is persoon.php: 

<?php include_once("persoon.class.php"); ?><!DOCTYPE html><html lang="nl">	<head>		<title>OOP in PHP</title>		<meta name="author" content="Robin Bosman">	</head>	<body>	<?php		$jan = new persoon("Jan", 18, "man");		$truus = new persoon("Truus", 20, "vrouw");		echo '<br />De leeftijd van ' . $truus->getNaam() . ' is: ' . $truus->getLeeftijd();		$truus->setLeeftijd(22);		echo '<br />De leeftijd van ' . $truus->getNaam() . ' is: ' . $truus->getLeeftijd();		$renee = new Persoon("Renee", 21, "man");		$renee->setGeslacht("vrouw");		echo '<br />Het geslacht van ' . $renee->getNaam() . ' is: ' . $renee->getGeslacht();		$thara = new Persoon("Thara", 19, "vrouw");		$thara->setLeeftijd(20);		echo '<br />De leeftijd van ' . $jan->getNaam() . ' is ' . $jan->getLeeftijd();				echo '<br />De leeftijd van ' . $thara->getNaam() . ' is ' . $thara->getLeeftijd();		printGegevens($thara);		printGegevens($jan);		unset($jan);	?>	</body></html>

this is : persoon.class.php: 

<?phpclass Persoon{	private $_naam;	private $_leeftijd;	private $_geslacht;	function __construct($persoonsnaam, $leeftijd, $geslacht){		$this->_naam = $persoonsnaam;		$this->_leeftijd = $leeftijd;		$this->_geslacht = $geslacht;		echo '<br /> Nieuw Persoon object ' . $persoonsnaam . ' wordt aangemaakt.';	}	function setNaam($persoonsnaam){		if(is_string($persoonsnaam)){			$this->_naam = $persoonsnaam;		}		else{			echo '<br />Datatype error in setNaam() method';		}	}	function getNaam(){		return $this->_naam;	}	function setLeeftijd($leeftijd){		if(is_integer($leeftijd)){			$this->_leeftijd = $leeftijd;		}		else{			echo '<br /> Datatype error in setLeeftijd()';		}	}	function getLeeftijd(){		return $this->_leeftijd;	}	function setGeslacht($geslacht){		$this->_geslacht = $geslacht;	}	function getGeslacht(){		return $this->_geslacht;	}		public function printGegevens(){		$gegevens = '<br />de gegevens van ' . $this->_naam . ' zijn: </br>Leeftijd: ' . $this->_leeftijd . '<br />Geslacht: ' . $this->_geslacht;		echo $gegevens;	}	function __destruct(){		echo '<br />Persoon object ' . $this->_naam . ' wordt verwijderd';	}	}?>

Thanks for helping me out :) 

Beneath this mask there is an idea, and ideas are bulletproof.

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Posted

off topic - i read that title as pulp fiction, i'll go away now.

 

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Posted
  • Solution

I don't know PHP, but aren't you supposed to do it like this?

object->function()

Its a function of an object, like getNaam(). The error is occuring because it assumes you made a printGegevens() function, but you haven't. The printGegevens() is supplied under an object, but you are not using the correct syntax to use that function.

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I don't know PHP, but aren't you supposed to do it like this?

object->function()

Its a function of an object, like getNaam().

thanks , I think I need a break now what I was doing is fucked up. 

Beneath this mask there is an idea, and ideas are bulletproof.

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