I am doing an ajax call to a PHP which should do 2 SQL queries. The queries look like this:
$sql = "UPDATE customers SET customers_newsletter=1 WHERE customers_id ='".$cid."'";
$sql .= "INSERT INTO coupons (coupon_id,
coupon_type,
coupon_code,
coupon_amount,
coupon_minimum_order,
coupon_start_date,
coupon_expire_date,
uses_per_coupon,
uses_per_user,
coupon_active)
VALUES ('".$cid."',
'NL_".$cid_substr."".$cid."',
'F',
'5.0000',
'100.0000',
'".date("Y-m-d H:i:s")."',
'".$expiredate."',
'1',
'1',
'Y'
)";
mysqli_multi_query($con,$sql);
In another php file the exact same code already worked, i there copied an sql entry to another table and then deleted it from the current one.
If i do only one of the queries it works, but i need to get them to work together.
Any ideas why it is not working?
UPDATE:
I now followed the link for preventing sql injection in the comment and i got the following code now:
<?php
$mysqli = new mysqli("server", "user", "pw", "db");
// TODO - Check that connection was successful.
$unsafe_variable = $_GET['cid'];
$stmt = $mysqli->prepare("INSERT INTO coupons (coupon_id) VALUES (?)");
// TODO check that $stmt creation succeeded
// "s" means the database expects a string
$stmt->bind_param("s", $unsafe_variable);
$stmt->execute();
$stmt->close();
$mysqli->close();
mysqli_close($con);
?>
It is still not working. Where is the fault?
What you're currently running is the same as:
$sql = "UPDATE customers SET customers_newsletter=1 WHERE customers_id ='".$cid."' INSERT INTO coupons (coupon_id,
coupon_type,
coupon_code,
coupon_amount,
coupon_minimum_order,
coupon_start_date,
coupon_expire_date,
uses_per_coupon,
uses_per_user,
coupon_active)
VALUES ('".$cid."',
'NL_".$cid_substr."".$cid."',
'F',
'5.0000',
'100.0000',
'".date("Y-m-d H:i:s")."',
'".$expiredate."',
'1',
'1',
'Y'
)";
mysqli_multi_query($con,$sql);
Which, if you notice right after the first query it starts right into the INSERT. If you ran this in anything that would give you the SQL error (or echo'd the sql error here) you'd likely see that there is a syntax error because the UPDATE query is never closed. Try adding a ;
to the end of the update statement, like so:
$sql = "UPDATE customers SET customers_newsletter=1 WHERE customers_id ='".$cid."';";