Ex:
$stmt = $db->prepare("SELECT * FROM BD");
$stmt->execute();
Why does not it work if I instantiate the execute method of the same instance? since the value has already been passed to the prepare method? ex:
$db->prepare("SELECT * FROM DB");
$db->execute();
The% w_that is your PDO class has the prepare , which is responsible for preparing an execution command (SQL ) and that it returns a PDOStatement a> and it has execute method, that is, $db does not have this $db method, but has a method that returns a PDOStatement that has the execute.
In execute you can directly use the PDO :: query , which also returns a PDOStatement already run being the one with data return and PDO::exec , which executes Insert, Update, and Delete commands that return the number of affected rows.
//select
$db->query("SELECT * FROM BD");
//insert, update e delete
$rowsafetados = $db->exec("INSERT INTO BD ...");
There is separation of responsibility between the classes ( POO ) and each class has its own responsibility, $db connection and fast methods, and PDOStatement that uses $db as a connection to its methods.
Because $db is an instance of the class PDO and it represents a < strong> connection with the database. $stmt in turn is an instance of the class PDOStatement , which in simple terms represents a single query .
Just think a little: $db can not be executed because it is a connection, running a connection would be the act of connecting, but you are already connected. So $stmt represents a query , and it's who you run .
Still the class PDO has a function PDO::exec which allows you to perform queries quickly, but without the possibility to search results, returning only the number of modified lines .