RE: [PHP] Problem with MySQL Query

2003-07-23 Thread Chris Kay
IL PROTECTED] Sent: Thursday, 24 July 2003 8:24 AM To: David Nicholson Cc: PHP List Subject: Re: [PHP] Problem with MySQL Query Tried NOT LIKE and that didnt exclude it either. I am trying to exclude only 'Meal Plans' Phil - Original Message - From: "David Nicholson" <

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread Phillip Blancher
dnesday, July 23, 2003 6:19 PM Subject: Re: [PHP] Problem with MySQL Query > Hello, > > This is a reply to an e-mail that you wrote on Wed, 23 Jul 2003 at > 22:54, lines prefixed by '>' were originally written by you. > > > > $dbqueryshipping1 = "s

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread David Nicholson
Hello, This is a reply to an e-mail that you wrote on Wed, 23 Jul 2003 at 22:54, lines prefixed by '>' were originally written by you. > > $dbqueryshipping1 = "select * from tempuserpurchase where > > (usersessionid="$User_Shopping_Id") and day="1" and > > type<>'Meal Plans' > Tryed both me

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread Liam Gibbs
> > $dbqueryshipping1 = "select * from tempuserpurchase where > > (usersessionid=\"$User_Shopping_Id\") and day=\"1\" and > > type<>'Meal Plans' > > > > Tryed both methods and it is still not excluding anything matching Meal > Plans Been a short while since I used SQL with my PHP, but try pu

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread Phillip Blancher
> > $dbqueryshipping1 = "select * from tempuserpurchase where > (usersessionid=\"$User_Shopping_Id\") and day=\"1\" and > type<>'Meal Plans' > Tryed both methods and it is still not excluding anything matching Meal Plans --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread David Nicholson
Hello, This is a reply to an e-mail that you wrote on Wed, 23 Jul 2003 at 22:28, lines prefixed by '>' were originally written by you. > As per your suggestion > $dbqueryshipping1 = "select * from tempuserpurchase where > (usersessionid="$User_Shopping_Id") and day="1" and type!='Meal > Plan

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread Phillip Blancher
As per your suggestion $dbqueryshipping1 = "select * from tempuserpurchase where (usersessionid=\"$User_Shopping_Id\") and day=\"1\" and type!='Meal Plans' "; $resultshipping1 = mysql_db_query($dbname,$dbqueryshipping1); if(mysql_error()!=""){echo mysql_error();} $result1 = my

Re: [PHP] Problem with MySQL Query

2003-07-23 Thread CPT John W. Holmes
Take out the plus sign... type != 'Meal Plans' And using single quotes in your query might make things easier (no escaping...) $dbqueryshipping1 = "select * from tempuserpurchase where usersessionid='$User_Shopping_Id' and day='1' and type!='Meal Plans'"; You don't need quotes around '1' since i