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
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 with the database an
