Re: SQL help

That works well, Thanks all. "Warren Vail" <Warren.Vail@schwab.com> wrote in message news:72138202E59CD6118E960002A52CD9D20F5D4650@n1025smx.nt.schwab.com... > The problem is you are using a hard join which will only return rows where > there are matching entries in both tables. What you probably want is called > a left join, however you have a further complication. > > You have to have an image row to have a image_data.Default_Img = 'Yes'. So > this query is really designed to only return users who have default images > established (other images don't count either). > > Haven't tested this, but it reflects my understanding of your expectation > (haven't double checked syntax either, but you can at > http://www.mysql.com/doc/en/index.html); > > SELECT * FROM user_data > left join image_data > on user_data.User_ID = image_data.User_ID > where image_data.Default_Img in ('Yes', '') > GROUP BY user_data.User_ID > > hope this works for you, good luck, > > Warren Vail > > > -----Original Message----- > From: Marc Greenstock [mailto:marc@shockmedia.com.au] > Sent: Tuesday, February 10, 2004 6:53 PM > To: php-general@lists.php.net > Subject: [PHP] SQL help > > > Hi all here's my problem, > > I have two tables: 'user_data' and 'image_data' > > obviously user_data has all the user information and image_data has all the > image information > > in image_data are the fields: Image_ID int(11), User_ID int(11) and > Default_Img enum('Yes','No'). Users can have as many images as they like, > but it is optional. > > I want to select from user_data and image_data and get the results based > upon the rest of the query. The problem is my sql is only pulling out the > users with images and ignoring the rest > > my sql looks like this: > > SELECT * FROM user_data, image_data > WHERE user_data.User_ID = image_data.User_ID > AND image_data.Default_Img = 'Yes' > GROUP BY user_data.User_ID > > thanks for your help. > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php