mysqli_error() expects exactly 1 parameter, 0 given [duplicate]

Answer 1

Use mysqli_error($result) as mysqli_error expects the connection to be passed as a parameter.

Answer 2

As far as the sql error is concerned, does 'user' have permissions to alter the table?

Answer 3

You are mixing the object-oriented and the procedural styles of the mysqli API :

You are using object-oriented :

$result = new mysqli('localhost', 'user', 'password', 'db');

And, then, procedural :

echo "Error with MySQL Query: ".mysqli_error();

You should use either OO, or procedural -- but not both ; and if you choose procedural, the functions expect the link identifier passed as a parameter.

For instance, mysqli_error should be called either using the object-oriented API :

$link = new mysqli(...);
echo $link->error;

Or the procedural API :

$link = mysqli_connect(...);
echo mysqli_error($link);

(Of course, it will not change the fact that you are having an error in your SQL query, but it'll allow you to get the error message, which should help finding the cause of that error)