Insert auto-incremented ID using prepared statements

问题

When inserting a new record into a table with an auto-incrementing ID column, it is normally enough to give the ID field the value NULL or omit it from the INSERT query, as explained at How to insert new auto increment ID

INSERT INTO  `database`.`table` (`id`, `user`, `result`) VALUES (NULL, 'Alice', 'green')");

or

INSERT INTO  `database`.`table` (`user`, `result`) VALUES ('Alice', 'green')");

My question is - how do you do the same thing when using prepared statements. I have tried the following, using NULL:

$stmt = $db->prepare("INSERT INTO `db` (id, name, password, text) VALUES (NULL, ?, ?, ?)");
$stmt->bind_param('sss', $name, $password, $text);
$stmt->execute();

and the fowllowing, omitting the ID field:

$stmt = $db->prepare("INSERT INTO `test_db` (name, password, text) VALUES (?, ?, ?)");
$stmt->bind_param('sss', $name, $password, $text);
$stmt->execute();

When I run this I get nothing inserted and no error message in the browser. I think it is because it is trying to insert a duplicate value for the ID field (stackoverflow.com/questions/12179770/…) - but why it should do that when this seems equivalent to the non-prepared-statement way of inserting data, and then give no message, I'm not sure.

Any ideas most welcome!

来源：https://stackoverflow.com/questions/17592545/insert-auto-incremented-id-using-prepared-statements