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:[EMAIL PROTECTED]
Sent: Tuesday, May 27, 2003 2:47 AM
To: Matthew Moldvan
Cc: S. Cole; [EMAIL PROTECTED]
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" <[EMAIL PROTECTED]>
>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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