PHP error mysql

toadstool

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 if (IN PHP) I call $cart->insert_Pcart($item1, $item2);
 
class Cart 

{

:

function insert_Pcart($item_name,$p_qty)

      {

$insert = "INSERT INTO Cart (pname, qty) 

VALUES ($_POST[$item_name]','$_POST[$p_qty]')"; 
 Line69=>     if (mysql_query($insert, $conn)) { $cartID = mysql_insert_id($conn);
 
} else          { die ("Error inserting CART FORM data: " .mysql_error()); };
 
//$p_name=($_REQUEST["p_name"]);

}

I get the error
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\phpdev\www\project_shopping_cart.php on line 69
Error inserting CART FORM data:

Can anyone explain what this error means or how I might fix it?

Markus

Quote:

Originally Posted by toadstool

Expand|Select|Wrap|Line Numbers

 if (IN PHP) I call $cart->insert_Pcart($item1, $item2);
 
class Cart 

{

:

function insert_Pcart($item_name,$p_qty)

      {

$insert = "INSERT INTO Cart (pname, qty) 

VALUES ($_POST[$item_name]','$_POST[$p_qty]')"; 
 Line69=>     if (mysql_query($insert, $conn)) { $cartID = mysql_insert_id($conn);
 
} else          { die ("Error inserting CART FORM data: " .mysql_error()); };
 
//$p_name=($_REQUEST["p_name"]);

}

I get the error
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\phpdev\www\project_shopping_cart.php on line 69
Error inserting CART FORM data:

Can anyone explain what this error means or how I might fix it?

It means you're trying to supply a MYSQL-Link resource that isn't valid, that is, your $conn variable isn't a valid resource. You don't seem to set it anywhere, so why pass it? You can remove this and try again. If no resource is provided, MySQL will take the current (or last opened, I forget) link.

Dormilich

Quote:

Originally Posted by toadstool

Can anyone explain what this error means or how I might fix it?

this means that the 2nd parameter ($conn) is of the wrong type (it must be a resource type)

did you define $conn in the function (otherwise it is NULL, which is not a resource type)? probably the best is omitting that parameter (except you have more than one connection open, in this case you application needs aredesign)

darn…

toadstool

Interestingly, this did not produce an error

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function connect_cart()

      { 

        $conn = @mysql_connect($this->host,$this->user,$this->pass);  

       if (!$conn) { die("Connection failed: " .mysql_error());}

       if (mysql_select_db("test", $conn)) { 

        } else                                { die ("Could not locate test database " .mysql_error());}

      }

toadstool

If I want to set a cookie from this wouldn't removing $conn affect this?
e.g.
$whatever = mysql_insert_id($conn);
setcookie("cookiename",$whatever);

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