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    <p>In PHP, a constructor does not return.</p> <p>So your <code>get</code> method returns a <code>one</code> object, the first time it's called, then a <code>mysqli</code> object. Probably not what you want.</p> <pre><code>if( self::$_db == NULL ) { return new self(); // Here you return an object of class one } else { return self::$_db; // Here you return an object of type mysqli } </code></pre> <p>If you want to return the <code>mysqli</code> object, you do not need a singleton, as there is no need to create an instance of an object that's only here to return an instance of another object.</p> <p>The registry pattern would be better in such a case.</p> <p>If you need to provide methods (a wrapper for your DB object), then create a real singleton.</p> <p><strong>EDIT</strong></p> <p>I checked your updated code. Now you return always the <code>mysqli</code> instance. But you do not need to instantiate your own object. That's completely useless...</p> <p>If you really want to go with your kind of pattern, as golden said, in your static instance, checks if <code>self::db</code> is <code>NULL</code>. If yes, creates the <code>mysqli</code> instance, and assign it to <code>self::db</code>. Then returns it.</p> <pre><code>public static getDatabaseInstance() { if( self::$_db == NULL ) { self::$_db = new mysqli( ... ); } return self::$_db; } </code></pre> <p>Also set the constructor private, so users won't be able to create useless instances of your class. Or better make it public and throw an exception:</p> <pre><code>public function __construct() { throw new Exception( 'This class is not supposed to be instantiated' ); } </code></pre>
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    1. COFrom the PHP documentation: `void __construct ([ mixed $args [, $... ]] )`. See the `void`? http://php.net/manual/en/language.oop5.decon.php
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    2. COIt means the return value is ignored. So even if you return something (another object), it won't modify the type of the object instantiated by `new`.
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    3. COSee my edit... : )
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