On Monday 28 January 2002 20:18, B. Verbeek wrote:
> I already run the query without the value of the primary-key field:
>   $q_orders_id2 = "INSERT INTO orders(orders_datum, orders_sessid)";
>   $q_orders_id2.= " VALUES ('$date', '$sid')";
>   $r_orders_id2 = mysql_query($q_orders_id2, $sqllink_id);
>   $orders_id = mysql_insert_id($sqllink_id);
> <<
> The primary-key field is: "orders_id" (not in above query).
> Only "orders_datum" and "orders_sessid" are sent with the query.
> When I call mysql_insert_id with tha vars: "$q_orders_id" or
> "$r_orders_id" I get an warning (Warning: Supplied argument
> is not a valid MySQL-Link resource in /home/httpd/blabla/file.php
> on line 91) but not the wanted id.
> In the manual it says that mysql_insert_id() is used with the
> linkidentifier ($sqllink_id ? ) but not how to define the identifier
> in mysql_query()...
> >>How do I define a linkidentifier for a specific call to
>   mysql_query() and then use mysql_insert_id() with that
>   linkidentifier of the query?

When you initiate the connection to the db using mysql_connect() or 
mysql_pconnect() it returns a MySQL link identifier. 

In most cases you do NOT need to specify the link identifier when using 
mysql_query(). If you're only dealing with one database throughout your 
script then there's no reason for you to specify the link identifier because 
PHP automatically uses the most recently used one. That is it automatically 
chooses the correct link identifier to use.

So in your case what you want would probably be something like this:

  $q_orders_id2  = "INSERT INTO orders(orders_datum, orders_sessid)";
  $q_orders_id2 .= " VALUES ('$date', '$sid')";
  # for debugging, you may want to echo $q_order_id2
  $r_orders_id2 = mysql_query($q_orders_id2);
  $orders_id = mysql_insert_id();

Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

I used to be Snow White, but I drifted.
                -- Mae West

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