Re: Object declaration

"Cory D. Wiles" <cwiles@xxxxxxxxxxxxxx> · Fri, 08 Oct 2004 09:50:14 -0500

The reason why that you get the error is because you didn't call the 
class. How I do my sql/class manipulation would be something like this:

class User {
var $dbRecords = array();//array that will hold db records

function User() {
...constructor
}//

}//end class

$sql = "SELECT * FROM user WHERE username = '".$_COOKIE["username"]."' 
LIMIT 1";

$user_res = mysql_query($sql) or die("Query failed : " . mysql_error());

$user = new User();//call User class

while ($user_row = mysql_fetch_row($user_res)) {
  $user->dbRecords[] = array(
  'Username'       => $user_row["username"],
  'Password'       => $user_row["password"],
  'Agent Id'       => $user_row["agent_id"],
  'Reference'      => $user_row["reference"],
  'Level'          => $user_row["level"],
  'Email Address'  => $user_row["email_address"],
  'Email Username' => $user_row["email_username"],
  'Email Password' => $user_row["email_password"]
  );//end array
}//end while

To access the information you just stored in your array:

foreach ($user->dbRecords as $records):
  print $records['Username']."<br>";
  print $records['Password']."<br>";
  print $records['Agent Id']."<br>";
  print $records['Reference']."<br>";
  print $records['Level']."<br>";
  print $records['Email Address']."<br>";
  print $records['Email Username']."<br>";
  print $records['Email Password']."<br>";
endforeach;

If you want to a complete example that I have done

then go to http://www.randomthoughtprocess.com/AdminArticle.phps for the 
class and http://www.randomthoughtprocess.com/admin_article.phps for the

display.  You don't have to use an array to store your db values. I do 
that so I can keep my class and display logic seperate.

Nunners wrote:

Hi all,

In the ongoing saga of moving the application I've developed from Apache to
IIS, I've found another problem, with objects this time.

I've got a standard file that is included as a header, which includes the
following lines:

[script]

$query = "SELECT * FROM user where username='".$_COOKIE["username"]."' LIMIT

1"; 
$user_res=mysql_query($query) or die("Query failed : " . mysql_error()); 
while ($user_row=mysql_fetch_row($user_res)) {

	$user->username=$user_row["username"];

	$user->password=$user_row["password"];

	$user->agent_id=$user_row["agent_id"];

	$user->reference=$user_row["reference"];

	$user->level=$user_row["level"];

	$user->email_address=$user_row["email_address"];

	$user->email_username=$user_row["email_username"];

	$user->email_password=$user_row["email_password"];

}

[/script]

When I then try and recall the object variable $user->username it comes up
with an error:
Notice: Undefined variable: user in C:\www\portal\global\var_setup.php on
line 15

Notice: Trying to get property of non-object in
C:\www\portal\global\var_setup.php on line 15

What do I need to do to declare the object before I write sections to it?

Cheers
Nunners

--
Cory Wiles
Systems Engineer
ePerformax - Memphis, TN
901.751.4902
http://www.eperformax.com

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