From: James McConnell <[EMAIL PROTECTED]>
To: "Barry ." <[EMAIL PROTECTED]>, MySQL Mailing List <[EMAIL PROTECTED]>
Subject: Re: Learner Here Getting Frustraighted
Date: Thu, 08 Apr 2004 09:34:36 -0500
On 4/8/04 9:18 AM, "Barry ." <[EMAIL PROTECTED]> wrote:
> Hi
>
> I recently began to work through the Book entitled PHP & MySQL For Dummies
> and i am currently stuck towards the end of the forth chapter ive done
> everything as said so in the book but i keep getting an error message. Here
> is an outline of the problem:
>
> i have 2 tables one pet containing petName and petType other table is color
> containgin petName and petColor. The code which i keep getting errors on is:
>
> Select * from pet outer join color using (pet.petName=petcolor.petName) ;
>
> Now Ive posted my problem on several forums on the internet and have a wide
> range to replys suggesting what to do but nothing seems to work.
>
> Can anyone please help me??
>
> Barry Smith
>
> _________________________________________________________________
> Find a cheaper internet access deal - choose one to suit you.
> http://www.msn.co.uk/internetaccess
>
I think the reason that's not working is because that doesn't look like MySQL syntax for a join statement. From the MySQL documentation, try this:
SELECT * FROM pet OUTER JOIN color ON pet.petName=petColor,petName;
Keep in mind this will only return rows in pet that have corresponding rows in color. I hope that's what you're looking for.
The above code is not tested, but that's the proper JOIN syntax in MySQL.
Going through the documentation, I donšt see a USING keyword anywhere. That
looks suspiciously like Oracle code.
Hope that help!
_________________________________________________________________
Sign-up for a FREE BT Broadband connection today! http://www.msn.co.uk/specials/btbroadband
-- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
