Hey all,
I'm not sure if this is a trivial questions but in a PHP class:
MyClass:
class MyClass {
public $var1;
public $var2;
constructor() { ... }
public method1 () {
// Dynamically create an instance variable
$this->var3 = "test"; // Public....?
}
}
Main:
$test = new MyClass();
$test->method1();
echo $test->var3; // Would return "test"
Does this work?? How would I get this to work? Ps. I wrote this quickly so please disregard any errors I made with setting up the class or calling methods!
EDIT What about making these instance variables that I create private??
EDIT 2 Thanks all for responding - Everyone is right - I should have just tested it out myself, but I had an exam the next morning and had this thought while studying that I wanted to check to see if it worked. People keep suggesting that its bad OOP - maybe but it does allow for some elegant code. Let me explain it a bit and see if you still think so. Here's what I came up with:
//PHP User Model:
class User {
constructor() { ... }
public static find($uid) {
$db->connect(); // Connect to the database
$sql = "SELECT STATEMENT ...WHERE id=$uid LIMIT 1;";
$result = $db->query($sql); // Returns an associative array
$user = new User();
foreach ($result as $key=>$value)
$user->$$key = $value; //Creates a public variable of the key and sets it to value
$db->disconnect();
}
}
//PHP Controller:
function findUser($id) {
$User = User::find($id);
echo $User->name;
echo $User->phone;
//etc...
}
I could have just put it in an associative array but I can never correctly name that array something meaningful (ie. $user->data['name'] ... ugly.) Either way you have to know what is in the database so I do not really understand what the argument is that its confusing, especially since you can just var dump objects for debugging.
Thanks, Matt Mueller