Hello
I have a "base class" for my application, that holds a protected
variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work
(Call to a member function x on a non-object). How would I use the base
class's $db variable in it? 6 1229
Sorry, I forgot to mention that the class that extends the base class is
created in the base class.
Sweetiecakes wrote:
Hello
I have a "base class" for my application, that holds a protected
variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work
(Call to a member function x on a non-object). How would I use the base
class's $db variable in it?
Sweetiecakes <x@x.comwrote in comp.lang.php:
Hello
I have a "base class" for my application, that holds a protected
variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work
(Call to a member function x on a non-object). How would I use the base
class's $db variable in it?
Look up the scope resolution operator (::) to refer to properties/methods
of objects. Probably you want:
baseClassObj::db http://us3.php.net/manual/en/languag...ekudotayim.php
Hi
If I try to use "parent::$db", the following error shows up:
Access to undeclared static property
If I try to use "parent::db", the following error showsup:
Undefined class constant 'db'
Five By Five wrote:
Sweetiecakes <x@x.comwrote in comp.lang.php:
>Hello
I have a "base class" for my application, that holds a protected variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work (Call to a member function x on a non-object). How would I use the base class's $db variable in it?
Look up the scope resolution operator (::) to refer to properties/methods
of objects. Probably you want:
baseClassObj::db
http://us3.php.net/manual/en/languag...ekudotayim.php
..oO(Sweetiecakes)
>I have a "base class" for my application, that holds a protected variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work (Call to a member function x on a non-object). How would I use the base class's $db variable in it?
You do it with $this->db. Can you post some code?
Micha
Sweetiecakes <x@x.comwrote in comp.lang.php:
Hi
If I try to use "parent::$db", the following error shows up:
Access to undeclared static property
If I try to use "parent::db", the following error showsup:
Undefined class constant 'db'
Try parent->db or possibly $this->db. Without code posted, it's tought to
tell.
Top-posting is frowned on, by the way.
>
Five By Five wrote:
>Sweetiecakes <x@x.comwrote in comp.lang.php:
>>Hello
I have a "base class" for my application, that holds a protected variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work (Call to a member function x on a non-object). How would I use the base class's $db variable in it?
Look up the scope resolution operator (::) to refer to properties/methods of objects. Probably you want:
baseClassObj::db
http://us3.php.net/manual/en/languag...ekudotayim.php
On Aug 24, 4:13*pm, Sweetiecakes <x...@x.comwrote:
Sorry, I forgot to mention that the class that extends the base class is
created in the base class.
Sweetiecakes wrote:
Hello
I have a "base class" for my application, that holds a protected
variable called $db. $db is an ADODB object for the database.
When I make a class that extends the base class, $this->db does not work
(Call to a member function x on a non-object). How would I use the base
class's $db variable in it?
You're trying to define a class within a class? As far as I'm aware
that's not possible in PHP.
class foo
{
protected $db
class bar
{
if ($this -db)
{
...
}
}
}
is not possible in PHP as far as I'm aware. The following should be
valid (not tested).
class Foo
{
protected $db = 'test';
}
class Bar extends Foo
{
function getDb ()
{
return ($this -db);
}
}
$thisObj = new Bar;
echo ($thisObj -getDb ()); This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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