Re: Problem with MySQL Query
| From: | Phillip Blancher | Date: | Wed, 23 Jul 2003 21:27:03 +0000 |
| Subject: | Re: Problem with MySQL Query | ||
| References: | 1 2 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-156539@lists.php.net to get a copy of this message | ||
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();}
$result1 = mysql_num_rows($resultshipping1);
It is still not excluding "Meal Plans" from the listing.
> 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 it's an integer, but MySQL is
> forgiving.
>
> ---John Holmes...
>
> ----- Original Message -----
> From: "Phillip Blancher" <phil@ontarioweb.ca>
> To: "PHP List" <php-general@lists.php.net>
> Sent: Wednesday, July 23, 2003 5:08 PM
> Subject: [PHP] Problem with MySQL Query
>
>
> Problem with mySQL Query
> This is the query I have:
>
> $dbqueryshipping1 = "select * from tempuserpurchase where
> (usersessionid=\"$User_Shopping_Id\") and day=\"1\" and
type!=\"Meal+Plans\"
> ";
>
> What I want to do is to select everything from tempuserpurchase that
matchs
> the user session and the day, so long as type is not "Meal Plans"
>
> However it is not excluding "Meal Plans"
>
> Any Suggestions.
>
> Thanks in advance,
>
> Phil
>
>
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