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sketchMedia · 12-17-2008, 09:08 PM

Like the latter posts have stated, $member isnt defined anywhere in the scope of the called method, hence the notice about an undefined variable 'member'.
Also remove the single quotes from around $variable otherwise PHP will interpret that statement to be thus:

'Access the array element $variable within the $member array' and not 'Access the array element msn within the $member array' etc as you were intending.

This is because in PHP a variable name witin single quotes will not be expanded out and as a result will be treated as a normal string.
However if you were to use double quotes PHP would interpolate variable values in place of the variable names but in this case neither is needed, the variable name will do fine:

PHP Code:


			
$member[$variable];

Heres just a breif idea of what i think you are trying to achieve:

PHP Code:


			
class member
{
    private $member = array('msn' => 'msn@contact.com', 'other' => 'test@test.com');
    public function user($variable)
    {
        return $variable;
    }
    public function contact($variable)
    {
        return $this->member[$variable];
    }
}
$member = new member();

echo $member->contact('msn');

Obviously you will need to change the hard coded array property to be something dynamic (i.e. from the DB) but this should show you enough information to get you started.

Hope that helps.

12-17-2008, 09:08 PM	#5 (permalink)
sketchMedia The Prestige Join Date: Oct 2007 Location: Manchester, UK Posts: 854 Thanks: 32	Like the latter posts have stated, $member isnt defined anywhere in the scope of the called method, hence the notice about an undefined variable 'member'. Also remove the single quotes from around $variable otherwise PHP will interpret that statement to be thus: 'Access the array element $variable* within the $member* array' and not 'Access the array element msn* within the $member* array' etc as you were intending. This is because in PHP a variable name witin single quotes will not be expanded out and as a result will be treated as a normal string. However if you were to use double quotes PHP would interpolate variable values in place of the variable names but in this case neither is needed, the variable name will do fine: PHP Code: `$member[$variable];` Heres just a breif idea of what i think you are trying to achieve: PHP Code: `class member { private $member = array('msn' => 'msn@contact.com', 'other' => 'test@test.com'); public function user($variable) { return $variable; } public function contact($variable) { return $this->member[$variable]; } } $member = new member(); echo $member->contact('msn');` Obviously you will need to change the hard coded array property to be something dynamic (i.e. from the DB) but this should show you enough information to get you started. Hope that helps. __________________ mysql> SELECT * FROM `users` WHERE `users`.`clue` > 0; Empty set (0.00 sec)