This should working, right???

JakesSA

To me, this simple little class should work, but for some reason its not pulling the data out the db?

  class article
   {
   var $db_user = "xxxx";
   var $db_pass = "yyyyy";
   var $db_server = "localhost";
   var $db_name = "xyxyxyxyx"; 
   var $article_id;
   var $article_name ;
   function setArticle_id($i)
	{
	$this->article_id= $i;
	}
      function showArticle($article_id)
	    {
		if(!($connect = @ mysql_connect($this->db_server, $this->db_user, $this->db_pass))) die(mysql_error());
         if(!mysql_select_db($this->db_name, $connect))  die(mysql_error());
           $query = "SELECT article_name FROM articles WHERE article_id=".$this->article_id;
		   $result = mysql_query($query, $connect);
		   list($article_id, $article_name) = mysql_fetch_array($result);
		   echo  "Article Name: ".$this->article_name;	
		}
	}
$obj1 = new article();
$obj1->setArticle_id(1);
$obj1->showArticle();

mogster

Hmmm... shouldn't it be:
if(!($connect = @ mysql_connect($this->db_server, $this->db_user, $this->db_pass))) OR die(mysql_error());

instead of
if(!($connect = @ mysql_connect($this->db_server, $this->db_user, $this->db_pass))) die(mysql_error());

I'm not sure 😉

knutm

JakesSA

No, thats not it....you dont need the OR.

I ran the code through a normal function and it works great, but as soon as I stick it in a object....nothing gets outputed.

Im not to sure where the fault lies.

sgoku01

Does anything gets outputted by the function showArticle()??
If not, place a simple echo command at top of the function, and then check again.

And by the way: you're function showArticle() requires a variable, but you're not sending one...

JakesSA

showArticle() refers to itself with $this->; a member variable of the class.

All Im doing is defining the id with the setArticle_id() function and then wanting to pass it through showArticle() to get the name of the article, but still no such luck.

Thanks