Stumped on a aspect of references

[ScottyDNA]

Hi all,
I have the following two classes:

class Database
{
   var $conn;
   var $result;
   var $number;

   function conndb()
   {
      $this->conn = mysql_connect('localhost', 'user', 'password');
      mysql_select_db('database');
   }

   function query($sql)
   {
      $this->result = mysql_query($sql);
      $this->number = 1;
   }
}

class Test
{
   function test_ref($sql)
   {
      $test = new Database();
      $test->query($sql);
      $test->number = 6;
   }
}

Here is a copy of the program itself:

$db = new Database;
$db->conndb();
$query = "select * from table";
$db->query($query);
$num_result = $db->number;

$ref_test = new Test;
$query = "select * from table";
$ref_test->test_ref($query);
$num_result = $db->number;
mysql_close();
echo 'completed test';

A new database object is created when the test_ref() method is executed of the Test class. How do I modify my code so that the test_ref() method uses the already created Database object? I did do a search and found two threads with basically the same question, but I could not get either solution to work. I've been coding in PHP for a couple years but just recently decided to try the object oriented approach. The above code is just me trying to see what I can do (playing around). :p

Thanks in advance,
Scott

[sarahk]

two ways

create a variable in the class to hold the object and pass that in

or

make the database object global and in the class call the global variable again. Basically if you want a variable to be global you have to declare it as global when ever you want to use it.

does this make sense?

[milind24]

Try this may be this is your solution :

Yes by using session variable , insert this code in each and evrey page where u are using this object

<?
	// include calss file here

            session_start();
if(!isset($dbclass))
{
	$dbclass = new Database;
	session_register("dbclass");
}

//use of variable conn object as 
$dbclass->conn;

?>

[ScottyDNA]

Thank you all for your responses. SaraK, I am interested in the first way you mentioned and modified my Test class to the following.

class Test
{
   var $db;

   function Test($dbase)
   {
      $this->db = $dbase;
   }

   function test_ref($sql)
   {
      $db->query($sql);
      $db->number = 6;
   }
}

and here is the modified program.

$db = new Database;
$db->conndb();
$query = "select * from table";
$db->query($query);
$num_result = $db->number;

$ref_test = new Test($db);
$query = "select * from table";
$ref_test->test_ref($query);
$num_result = $db->number;
mysql_close();
echo 'completed test';

When it gets to $db->query($sql); in the test_ref() method, it gives the error "Call to a member function on a non-object". Shouldn't $db be the database object since I passed $db to the Test object or am I missing something? I will try making the $db variable global and see if that works. I now curious as to why the above will not work.

Thanks again,
Scott

[ScottyDNA]

I tried using the global variable idea but still get the "Call to a member function on a non-object" error. Here is the modified code. The Database class is still the same as it was in the first post. I did not try the session idea but assume it would not work if a global variable does not work.

Thanks again,
Scott

class Test
{
   function test_ref($sql)
   {
      $db->query($sql);
      $db->number = 6;
   }
}

global $db;
$db = new Database;
$db->conndb();
$query = "select * from table";
$db->query($query);
$num_result = $db->number;

$ref_test = new Test;
$query = "select * from table";
$ref_test->test_ref($query);
$num_result = $db->number;
mysql_close();
echo 'completed test';

[sarahk]

Just back from the beach! The America's Cup yachts are racing and it's a beautiful Auckland day!

Lets go back to "Plan A"

this is untested but the concepts will be right 🙂

class Test
{
   var $db;

   function Test($dbase)
   {
      $this->db = $dbase;
   }

   function test_ref($sql)
   {
      //$db = $this->db;

  //$db->query($sql);
  //$db->number = 6;

// I saved a lovely example of this syntax but can't find it. I'm pretty certain it's as basic as this
  $this->db->query($sql);
  $this->db->number = 6;

  return $this->db->number;
   }
}// End of Class "Test"

$db = new Database;
$db->conndb();
$query = "select * from table";
$db->query($query);
$num_result = $db->number;

$ref_test = new Test($db);
$query = "select * from table";
$num_result = $ref_test->test_ref($query);

mysql_close();
echo 'completed test';

ofcourse you will need to think about the value in passing the $db variable versus just letting the class create it's own.

Don't forget that sessions can now store objects so you don't need to close them between pages.

[volleytotal_ch]

Just back from the beach! The America's Cup yachts are racing and it's a beautiful Auckland day!

Alinghi will win anyway 😉

[sarahk]

staying off topic 😉 hope so! The crew that I have met to date are charming

[volleytotal_ch]

lol r u not intersted in sailing? 😉 just the crew ....

[sarahk]

thought I might learn to sail little boats next year.

And given that the teams are commercial and not really national I don't feel any loyalty to a team based on anything other than their behaviour. Sail well, act with honour etc and we'll love you.

There's been loads of rubbish in the press about the New Zealanders who are working for the other teams. Most of the real people think it's great, there's no way we could get the work locally for all these guys so they'd have to get days jobs like you and me.

Just one thing about Alinghi winning, it's just the Louis Vuitton that they're allowed to win. We still want Team New Zealand to win the America's Cup so we can have all the fun and games at home next time. Auckland has become a really fun place through hosting the Cup.

[volleytotal_ch]

i wish i could be there .... 🙁

may the better team win the America's Cup 😉