Error passing mysql connection as parameter - php

Question

I'm trying to build a basic API with PHP and mysql and depending on the url path, different database tables are used and therefore the connection needs to be made. But I keep getting this error:
Fatal error: Call to a member function prepare() on a non-object
in....line 6
dashboard.class.php:
class dashboard {
public function getData($conn) {
// Get latest status
$stmt = $conn->prepare("SELECT status FROM status_archive ORDER BY datetime DESC LIMIT 1 ");
$stmt->execute(); //line 6
$stmt->bind_result($status);
$stmt->fetch();
($status == '1' ? $status = 'up' : $status = 'down');
$stmt->close();
return $status;
}
}
Function that creates the database connection:
function db_connection($type) {
$db = $type.'_db';
syslog(LOG_INFO, 'DB: '.$db);
// Check to see if a development or production server is being used
if (strpos(getenv('SERVER_SOFTWARE'), 'Development') === false) {
$conn = mysqli_connect(null,
getenv('PRODUCTION_DB_USERNAME'),
getenv('PRODUCTION_DB_PASSWORD'),
$db,
null,
getenv('PRODUCTION_CLOUD_SQL_INSTANCE'));
} else {
$conn = mysqli_connect(getenv('DEVELOPMENT_DB_HOST'),
getenv('DEVELOPMENT_DB_USERNAME'),
getenv('DEVELOPMENT_DB_PASSWORD'),
$db);
}
// Check if successful connection to database
if ($conn->connect_error) {
die("Could not connect to database: $conn->connect_error " .
"[$conn->connect_errno]");
}
return $conn;
}
This is the code at the end of the file that initiates everything:
$path_array = explode("/", parse_url($_SERVER["REQUEST_URI"], PHP_URL_PATH));
$conn = db_connection($path_array[3]);
include 'classes/dashboard.class.php';
$dashboard = new dashboard;
$results = $dashboard->getData($conn);
echo json_encode($results, JSON_PRETTY_PRINT);

Answer 1

mysqli_connect can to return falsy value.
You should use OOP style (new mysqli) or check $conn:
$conn = mysqli_connect('localhost', 'my_user', 'my_password', 'my_db');
if (!$conn) {
die('Connection error (' . mysqli_connect_errno() . ') '
. mysqli_connect_error());
}
php.net - See "Procedural style" example.

Answer 2

It turned out it was due to variable scope.
I ended up using PHP $GLOBALS to get it to work.