oskare100 Posted January 22, 2007 Share Posted January 22, 2007 Hello,$item_number is one value but the MySQL querey inserts another.[code=php:0]Here is the code:$item_number = $_POST['item_number'];$sql46="INSERT INTO $sales_tbl (user_id, pack_id, ebay_item_name, ebay_item_id, ebay_txn_id, shipped_marked, paid_marked, payment_status, payment_date, payment_gross, payment_tax, payment_currency, payment_type, paypal_txn_id, paypal_memo, auction_closing_date, auction_multi_item, received) VALUES('".$row['user_id']."', '".$ident_row['file_id']."', '$item_name', '$item_number', '$ebay_transaction_id', '$shipped_marked', '$paid_marked', '$payment_status', '$payment_date', '$mc_gross', '$tax', '$mc_currency', '$payment_type', '$txn_id', '$memo', '$auction_closing_date', '$auction_multi_item', NOW())";$result46=mysql_query($sql46) or die( mysql_error() );}//Send new purchase message to buyer$to = "me@me.com";$subject = 'subject';$message = "Item number: $item_number";$headers = 'From: sales@ventiero.com' . "\r\n" . 'Reply-To: sales@ventiero.com' . "\r\n" . 'X-Mailer: PHP/' . phpversion();mail($to, $subject, $message, $headers);[/code]$item_number = 110079832134, and that's also the item number that is emailed to be. But when I look at the database 2147483647 has been inserted instead!? And it always inserts 2147483647 instead of the actual number so please help me, what's wrong?? The database ebay_item_id is int(15).Thanks in advance,/Oskar R Quote Link to comment Share on other sites More sharing options...
Jessica Posted January 22, 2007 Share Posted January 22, 2007 Just on first glance I'd say the int(15) is too small. Make it a bigint or something higher. Sounds like you're out of range. Is the int(15) unsigned as well? Quote Link to comment Share on other sites More sharing options...
oskare100 Posted January 22, 2007 Author Share Posted January 22, 2007 Hello,Doesn't int(15) mean that it can be 15 numbers? The item_number is never more than 12 numbers. Is the int(15) unsigned as well? Could you please explain that? Sorry for not understanding, but I'm not that good at MySQL.Best RegardsOskar R Quote Link to comment Share on other sites More sharing options...
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