Program interface to MySQL from PHP ($$$)

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Read pages 361 to 365 about opening a DB and creating a table.

Notes:
  • the new mysqli_connect( ) creates a object that represents the connection, if called with the first three arguments only, it just connects to the mySQL server without a specific DB, if called with four arguments, you supply the DB name to use in the fourth argument. The first argument can be the value null to indicate that you are connecting to the server as localhost.
  • Here is example code to connect to the mySQL server and create a db named testDB:
<?
$conn = new mysqli( NULL, "username", "password"); // conn to mysql with localhost

if (mysqli_connect_errno()) {
printf("Connection failed: %s\n", mysqli_connect_error());
exit();
} else {
$sql = "CREATE DATABASE testDB;"
$result = mysql_query($conn, $sql);
if($result == TRUE) {
echo "DB: testDB has been created";
} else {
printf("DB create failed: %s\n", mysqli_error());
}
}
mysqli_close($conn);
?>

Question?

In the code on page 364, what value will $res have if the table could not be created because of an error?