在SQL语法和PHP中出错

Question

I am getting the following error and can't figure out what the problem is.

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order (orderid, customerid, productid, brand, model, price, amount, totalcost) V' at line 1

//connect to database
$connection = mysql_connect("localhost","root","") or die ("Can't connect");
mysql_select_db("shoppingcart", $connection) or die ("Can't connect");

//get order id
$vol = mysql_query("SELECT orderid FROM ordertracking WHERE email='$email'");
while($volume=mysql_fetch_array($vol)) {
  $orderid = $volume['orderid'];
}
echo $orderid;
// add new order
$order = "INSERT INTO order (orderid, customerid, productid, brand, model, price, amount, totalcost) VALUES ('$orderid', '$customerid', '$productid', '$brand' , '$model', '$price', '$amount', '$totalcost')";
if (!mysql_query($order,$connection)) {
  die('Error: ' . mysql_error());
  echo "Sorry, there was an error";
}
echo "New order added" . "<br />";
mysql_close($connection);

Answer 1

ORDER is a mysql resered word enclose it in backticks ``.

You should not have a table or a column name conflicting with mysql reserved words otherwise you must have to enclose those in backticks.

$order = "INSERT INTO `order` (orderid, customerid,...

Answer 2

Like this:

INSERT INTO `order` (...) (ALT Gr+7)

If this solves the problem, give the credits to Shakti Singh.

Answer 3

The keyword ORDER is a reserved keyword in sql. Means you cannot use it

so this will produce an error:

$order = "INSERT INTO order (orderid, customerid, productid, brand, model, price, amount, totalcost) VALUES ('$orderid', '$customerid', '$productid', '$brand' , '$model', '$price', '$amount', '$totalcost')";

Insert into ORDER

the order in the above statement should either be in backticks (as Mentioned by shakti) like

INSERT INTO 'order' (orderid,......

or you can enclose the reserved keywords in square brackets like

INSERT INTO [order] (orderid,......

for more information check this thread stack overflow question

Answer 4

something else is that any code after die will not work i mean :

if (!mysql_query($order,$connection)) {
 die('Error: ' . mysql_error());
 echo "Sorry, there was an error";}

this code

echo "Sorry, there was an error";

will not work.and using this code :

die('Error: ' . mysql_error());

is not recomended at all due to security reasons