James McConnell wrote:
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
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 don't see a USING keyword anywhere. That looks suspiciously like Oracle code.
Hope that help!
USING is documented in the manual <http://www.mysql.com/doc/en/JOIN.html>. USING expects a list of columns which exist in both tables. The following are quivalent:
SELECT * FROM pet JOIN color ON pet.petName = color.petName; SELECT * FROM pet JOIN color USING (petName);
Michael
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