Re: Double Trouble!

[Leif K-Brooks <eurleif@buyer-brokerage.com>]

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:  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.



--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php