PHP MySQL事务未插入所有查询

Question

I got a problem. I run a transaction containing 2 queries but only one query is writing on database.. The other is not. How is that possible ? I thought it was either both or none, that's the point of using a transaction right ?

Here is part of my code :

mysqli_autocommit($conn, FALSE);
mysqli_query($conn, "INSERT INTO my_table_exemple (orderId, packageName, productId, purchaseTime, purchaseState, purchaseToken) 
                             VALUES ('".$_POST["orderId"]."', '".$_POST["packageName"]."', '".$_POST["productId"]."', '".$_POST["purchaseTime"]."', ".$_POST["purchaseState"].", '".$_POST["purchaseToken"]."')");

mysqli_query($conn, "UPDATE user
                             SET end_date='".$calcEndDate."'
                             WHERE id=".$_POST['user_id']);

/* Validation de la transaction */
if (!mysqli_commit($conn)) {
    $row_array['status'] = "error";
    $row_array['msg'] = "Error: " . mysqli_error($conn);
} else {
    $row_array['status'] = "success";
    $row_array['msg'] = "Transaction success";
}   
echo json_encode($row_array);

This is returning transaction success, updating user table but no record inserted in my_table_exemple.

I have an error on my insert query I guess, I'm going to debug it but I don't want to go ahead without knowing why my transaction is not working the way it is supposed to.

Answer 1

You need to check after each query for any errors - on any error you would rollback, if no errors after all of your queries you would commit. Only the queries with no error will be committed & the commit reports any errors on the commit - not preceding queries.