RE: [PHP-DB] Double Trouble!

That makes sense ... guess I shouldn't respond to questions so late at night and with any bit of alcohol. :P S.: did you find a workaround? Regards, Matt. x116 -----Original Message----- From: Leif K-Brooks [mailto:eurleif@buyer-brokerage.com] Sent: Tuesday, May 27, 2003 2:47 AM To: Matthew Moldvan Cc: S. Cole; php-db@lists.php.net Subject: Re: [PHP-DB] Double Trouble! It's calling the mysql_query function that runs a query, not doing something with the result it generates. If doing something with a functions result called the function again, then function foo(){ echo "Foo called."; return 2; } $result = foo(); echo $result + 2; would result in "Foo called.Foo called.4" being outputted, instead of "Foo called.4". Matthew Moldvan wrote: >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: [PHP-DB] 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 >> >> >> > > > > -- The above message is encrypted with double rot13 encoding. Any unauthorized attempt to decrypt it will be prosecuted to the full extent of the law.