PHP Mysqli inserting a row to a table with auto_increment column

Question

Im working on a table that has 4 columns and the first one is an auto incrementing integer called id.

If im going to insert into this table using mysqli prepared statements I keep having trouble inserting a query that works. Using PHPmyAdmin It tells me to give it NULL. Ive tried this:

$query = "INSERT INTO tbl (id, col2, col3, col4) VALUES ('NULL', ?, ?, ?)";
$stmt -> bind_param("ssi", $col2, $col3, $col4)

and this

$query = "INSERT INTO tbl (id, col2, col3, col4) VALUES (NULL, ?, ?, ?)";
$stmt -> bind_param("ssi", $col2, $col3, $col4)

and only give bind_param 3 arguments (the last 3). Neither of those work. I also tried this:

$null = NULL;
$query = "INSERT INTO tbl (id, col2, col3, col4) VALUES (?, ?, ?, ?)";
$stmt -> bind_param("issi", $null, $col2, $col3, $col4);

None of these work. Is there a standardized way of inserting into this type of table?

Answer 1

Just skip the id field, MySQL will fill it automatically:

$query = "INSERT INTO tbl (col2, col3, col4) VALUES (?, ?, ?)";
$stmt->bind_param("ssi", $col2, $col3, $col4)

Answer 2

This one

$query = "INSERT INTO tbl (id, col2, col3, col4) VALUES (NULL, ?, ?, ?)";
$stmt -> bind_param("ssi", $col2, $col3, $col4)

should surely work. What is the exact error you get?

Now I look better, you have $query and $stmt. What do you have in between? Probably you are missing some part.

It should be

$stmt = $mysqli->prepare("INSERT INTO tbl (id, col2, col3, col4) VALUES (NULL, ?, ?, ?)");
$stmt -> bind_param("ssi", $col2, $col3, $col4);
$stmt ->execute();

Answer 3

This should work, because id is added automatically (incremented for this reason) by mysql:

 $query = "INSERT INTO tbl (col2, col3, col4) VALUES (?, ?, ?)";

In some cases, you have to insert auto_incremtnt field explicitly, if this is the case then you can use the INSERT IGNORE statement, see mysql manual for more info about it.

Answer 4

If the id field is the auto_increment, then just don't specify it in your insert query :

$query = "INSERT INTO tbl (col2, col3, col4) VALUES (?, ?, ?)";

And, of course, don't try to bind any parameter to it ;-)

As it's generated by MySQL, there is no need to pass that column.