Looks to me like you are calling the mysql_query() function twice here: > $result = mysql_query( $query, $link ); << first > if ( ! $result ) << second > die ( "Unable to Add Payment to Database: ".mysql_error() ); try this instead: > if ( !$result=mysql_query($query, $link) ) > die ( "Unable to Add Payment to Database: ".mysql_error() ); Let me know if that works or not ... it should, though. Regards, Matt. ----- Original Message ----- From: "S. Cole" <s.cole@roadrunner.nf.net> To: < > Sent: Sunday, May 25, 2003 4:04 PM Subject: Double Trouble! > I can't seem to find out why I am getting two copies of the same entry in my > database. > > =============================================================== > print "Pending Payment - Adding to Database<br>"; > > $query = "INSERT INTO payments (subscr_id, pmt_date, txn_id, > exchange_rate, > amt_paid, payment_status, pending_reason, payer_email, payer_id, > payer_status, verify_sign, paypal_fee) > > VALUES( '$subscr_id','$payment_date','$txn_id','$exchange_rate', > '$mc_gross','$payment_status','$pending_reason','$payer_email', > '$payer_id','$payer_status','$verify_sign','$mc_fee')"; > > $result = mysql_query( $query, $link ); > if ( ! $result ) > die ( "Unable to Add Payment to Database: ".mysql_error() ); > > print "Pending payment - Added to Database<br>"; > =============================================================== > > The print statements at the beginning and end only run once. > > The script is not called from anywhere else. I don't have any other scripts > running to insert the same info. When I "//" out the query, nothing is > added to the database. > > I have a very similar script for adding a customer to another datbase and > now it's doing the same thing. > Could it be possible that I change some sort of setting in mysql? > > S. Cole > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
