Is it possible to instantiate a function in php?

Question

I am looking a PHP script example that contains the mysqli extension. The code in question is as follows:

$conn = new mysqli("myServer", "myUser", "myPassword", "Northwind");

I'm confused as to wether this is a function or a class instance. If it is a function, is it possible to instantiate a function?

Answer 1

What confuses you is the constructor.

This is a special public function inside a class which get's called straight after an instance gets created.

mysqli in your case is a class, which has a constructor like this:

public function __construct($server, $user, $password, $dbname) {
 // do something
}

See:

http://php.net/manual/en/mysqli.construct.php

http://php.net/manual/en/language.oop5.decon.php

Geekfact:

You can somehow get an "instance" of a function by using closures. These arent real instances, just callables .

function createInstance() {
    return function($a, $b) {
        return $a + $b;
    };
}

$myIndependendFunction = createInstance();

echo $myIndependendFunction(3, 2); // prints 5

Answer 2

You declare functions and use it , you declare objects (mysqli object in this case) and you instantiate it .

You can read the manual. http://www.php.net/manual/en/language.oop5.basic.php

Answer 3

If you want to contain your instantiation logic in one place you could wrap your current code within a function and return the ready to use mysqli class?

Example:

<?php

function getMysqliConnection(){
    return new mysqli("myServer", "myUser", "myPassword", "Northwind");    
}

$conn = getMysqliConnection();