Wow, our posts crossed in the mail and you suggested the same approach that
I discovered independently!
(Great minds think alike, right?)
I'm working remotely, the server is many states away, and I don't have a
local developement environment. I'm modifying the actual working site,
though I try to
I'm going re-post this query. Roger Backlund had been attempting to help me
but I've either stumped him or he's busy having a life :) Besides, my
first posting had several problems, since I was working from memory.
Finally, in the process of preparing this query, I found a small change
that
> I'm going to send direct from the programming computer, using a different
> e-mail address.
Ok, I reply to the list, but CC to you. Think I found your problem...
> I did find that I had one invalid foreign key value in table A, but my
> problem remains...
>
> I tried this:
>
> Database PosenL
* =James Birkholz=
> In a message dated 1/12/02 10:48:45 AM Central Standard Time,
> [EMAIL PROTECTED] writes:
> ---snip---
> << SELECT A.Name, B.Name, P.ID
>FROM Persons P
>LEFT JOIN QualityA A USING(A_ID)
>LEFT JOIN QualityB B USING(B_ID)
>WHERE P.ID = thatGuy; >>
> ---snip---
>
In a message dated 1/12/02 10:48:45 AM Central Standard Time,
[EMAIL PROTECTED] writes:
---snip---
<< SELECT A.Name, B.Name, P.ID
FROM Persons P
LEFT JOIN QualityA A USING(A_ID)
LEFT JOIN QualityB B USING(B_ID)
WHERE P.ID = thatGuy; >>
---snip---
That doesn't work, get an error as it
James,
> I'm new to the list, to mysql and to dynamic website programming. I'm not
> new to programming, had my nose in Access97 for the last few years, off and
> on. So I'm used to being coddled with sql and can't find a syntax that
> works for this situation:
>
> (I'm using phpMyAdmin to work
