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Good morning Freaks, I hope you're all having a productive week. I've a question about an issue I'm having accessing static properties. I've been doing some reading on them but haven't yet deployed one successfully. I have a bunch of classes and I want to make the database query tables all static so they can be used cross-class, so to speak. Here's a sample attempt -> class Post { private $conn; private $user_obj; public static $table = "posts"; public function __construct($conn, $username) { $this->conn = $conn; $this->user_obj = new User($conn, $username); // $this->table = "posts"; } and accessing it here -> public function getBreakingNews() { $query = mysqli_query($this->conn, "SELECT * FROM self::$table WHERE type='breaking' ORDER BY RAND()"); This gets me an error that $table doesn't exist as does using Post::$table. When I assign $this->table = 'posts' (the commented out line in __construct) and access it by $this->table all works great. What am I doing wrong that it won't find that public static property? In the manual there's this example, which I feel like I've stayed true to -> class Foo { public static $my_static = 'foo'; public function staticValue() { return self::$my_static; } } ..... print Foo::$my_static . "\n"; The only difference I see is that they're printing the value and I'm trying to use it in a database query. Can anyone guide me through this mistake I'm apparently making and can't see? I've tried both self::$table and Post::$table, self::table and Post::table, every permutation I can think of but still the variable is saying it's unassigned. Share your knowledge please

Hi Freaks, I'm looking for advice if someones willing to give it. Here's the situation -> I've been working on a project, I started to learn PHP specifically to complete this idea I had. My code has evolved a lot over time as I've started understanding more. Up until today I've been working on it with just the registration functionality, no login. I had my username hardcoded into the $user_obj instantiation. I decided I wanted to try to make category subscription functionality and doing that I realized I was better off finishing the login form first so as to get a users subscriptions into a session variable at login. This has brought about the issue of getting an unassigned variable warning from the User class when not logged in. How I made all my other classes was putting a $user in the __construct parameter for each class. I now feel this may have been a rookie error since I'm having problems with error messages especially undefined array keys and variables when there isn't a session started. It's become a bit of a mess. So the advice I'm looking for and hoping to find here is how you folks handle non $_SESSION sessions, when a user is just scrolling the site not logged in. Did I make a mistake requiring $user for each class __construct? should I move the $user parameter to only the methods that require them? Is there a simpler solution that my inexperience causes to elude me? What would you folks do in this situation?

I am using parent::__construct() in almost every classes to connect Mysql DB. Example class class secondClass extends dbconnect { public function __construct() { parent::__construct(); dbconnect class class dbconnect { private $mysqli_handler; public function __construct() { try { mysqli_report(MYSQLI_REPORT_STRICT); $this->mysqli_handler = mysqli_connect(DB_HOSTNAME, DB_USERNAME, DB_PASSWORD, DB_DBNAME); } catch (mysqli_sql_exception $e) { throw new Exception('Error: Could not make a database link using ' . DB_USERNAME . '@' . DB_HOSTNAME . '!'); } if ($this->mysqli_handler->connect_error) { trigger_error('Error: Could not make a database link (' . $this->mysqli_handler->connect_errno . ') ' . $this->mysqli_handler->connect_error); } $this->mysqli_handler->query("SET NAMES 'utf8'"); $this->mysqli_handler->query("SET CHARACTER SET utf8"); $this->mysqli_handler->query("SET CHARACTER_SET_CONNECTION=utf8"); $this->mysqli_handler->query("SET SQL_MODE = ''"); $this->mysqli_handler->query("SET time_zone = 'Asia/Kolkata'"); } Is this create multiple instance of mysql dbconnection? I am frequently getting mysql connection error on my shared hosting. If so how to avoid? I am using PHP Version 7.4.16, some detailed explanation will be useful for me as I am using like this for many projects. Thank you for your time. Prabakaran