Prepared statement database connection must be instansiated first? - php

Question

The parts in bold are what I am questioning. Inside the search_for_new_user function, if I change $conn->prepare to $this->db_connection()->prepare. I receive a lost connection error. However in the function right above it db_conn_test I can use this syntax. In both cases I am returning the $connection so I don't understand why there must be a difference in syntax.
class Database {
function db_connection() {
$server = "localhost";
$user = "user";
$password = "password";
$database = "database";
return $connection = new mysqli($server, $user, $password, $database);
}
function db_conn_test() {
if (**$this->db_connection()->connect_errno**) {
die($this->db_connection()->connect_errno . ": " . $this->db_connection()->connect_error);
} else {
echo "connected to mysql database";
}
}
function search_for_new_user($email) {
**$conn = $this->db_connection();**
if ($stmt = **$conn->prepare**("SELECT email FROM users where email = ?")) {
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($result);
$stmt->fetch();
echo $result;
$stmt->close();
$conn->close();
}
}
}

Answer 1

In db_conn_test you call db_connection twice only if you got connection error during first db_connection call, so in this case connection to DB is not created.
But in search_for_new_user you create connection twice.
I.e.:
in db_conn_test:
// if connection not created, because you got error
if ($this->db_connection()->connect_errno) {
// therefore each time you call db_connection(),
// you again try create connection, and got same error
// and return it in die text
die($this->db_connection()->connect_errno . ": " . $this->db_connection()->connect_error);
} else {
echo "connected to mysql database";
}
but in search_for_new_user: you call db_connection() and create connection(if all is ok). And then if you call db_connection in second try, first connection is gone away and you got error.
Your class should looks like this:
class Database {
protected $connection;
function db_connection() {
if ($this->connection !== null) {
return $this->connection;
}
$server = "localhost";
$user = "user";
$password = "password";
$database = "database";
return $this->connection = new mysqli($server, $user, $password, $database);
}
}