<?php
include("db.php");
//an object of type db
$D=db_object();
//the class below
$C = new DbCreator($D);
//passing the $D reference
$C->CreatephpDatabaseObject("blah",$D);

class DbCreator{

public $D1;

function __construct($D)
{

$this->D1=$D;
}
function CreatephpDatabaseObject($database,$D1)
{
//WHY DOESN'T CODE COMPLETION WORK HERE
$this->D1->

}

function CreateClass($database)
{
echo "Class Db_".$database."(){";


}
}
?>

Ive updated the code so there are no errors but I still dont get code completion????
<?php
include("db.php");
$D = return_db_object();
$C = new DbCreator($D);
$C->CreatephpDatabaseObject("blah",$D);

class DbCreator{

public $D2;


public function CreatephpDatabaseObject($database,$D1)
{
$this->D2=$D1;

THIS CODE WORKS BU NO CODE COMPLETION!!!!
$data = $this->D2->SelectQuery("select * from urog_account");
$this->CreateClass($database);


}

public function CreateClass($database)
{
echo "Class Db_".$database."(){";


}
}
?>


Thanks

Got it!!

If you can't get the parser to see which type the variable has (it happens plenty since sometimes includes are dynamic etc), you can always use @var as a type indicator. It has helped me a lot in the past



Just a fictional example. If the parser didn't know what the type was, it knows now

Edit: The /** is not a typo, it needs to be 2 asterisks so that the block is interpreted as a DocBlock