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Home/ Questions/Q 6023497
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T04:00:50+00:00 2026-05-23T04:00:50+00:00

I have the following class: class DB { private $name; public function load($name) {

  • 0

I have the following class:

class DB {

    private $name;

    public function load($name) {
        $this->name = $name;
        return $this;
    }

    public function get() {
        return $this->name;
    }
}

At the moment if I do:

$db = new DB();
echo $db->load('foo')->get() . "<br>";
echo $db->load('fum')->get() . "<br>";

This outputs “foo” then “fum”.

However if I do this:

$db = new DB(); 
$foo = $db->load('foo');
$fum = $db->load('fum');
echo $foo->get() . "<br>";
echo $fum->get() . "<br>";

It always outputs “fum”.

I can sort of see why it would do that, but how could I keep the variable seperate to each instance without having to create a new instance of DB?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T04:00:51+00:00Added an answer on May 23, 2026 at 4:00 am

    IF you mean for DB to be some sort of database connectivity object (which I assume you are), multiple instances may not be the best choice. Perhaps something like this may be what you are trying to do:

    $db = new DB(); 
    $foo = $db->load('foo')->get();
    $fum = $db->load('fum')->get();
    echo $foo . "<br>";
    echo $fum . "<br>";
    

    If what I think you are trying to do is correct, it may be better to separate your get logic from the DB class into its own Record class.

    class Record {
        function __construct($name) {
            $this->name = $name;
        }
    
        function get(){
            return $this->name;
        }
    
        private $name;
    }
    
    class DB {
    
        public function load($name) {
            return new Record($name);
        }
    }
    
    $db = new DB(); 
    $foo = $db->load('foo');
    $fum = $db->load('fum');
    echo $foo->get() . "<br>";
    echo $fum->get() . "<br>";
    
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