Re: 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

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