I get the following error
Warning: mysqli_error() expects exactly 1 parameter, 0 given
The problem is with this line of the c
mysqli_error() needs you to pass the connection to the database as a parameter. Documentation here has some helpful examples:
http://php.net/manual/en/mysqli.error.php
Try altering your problem line like so and you should be in good shape:
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error($myConnection));