"Martijn Tonies" <[EMAIL PROTECTED]> wrote on 02/21/2005 10:20:28 AM:
>
>
> > Hi,
> > why not try:
> > SELECT COUNT(s.Id)+COUNT(se.Id)
> > FROM subs s
> > INNER JOIN subs_erased se ON s.Id=se.Id
> > WHERE s.Id=1;
> >
> > /Johan
>
> This won't return the same result if there's no entries
> in "s
Hi Martijn,
yes of course you're right but
SELECT COUNT(s.Id)+
(SELECT COUNT(se.Id) FROM subs_erased se WHERE s.Id=se.Id)
FROM subs s
WHERE s.Id=1
might work, at least in 4.1.x. I did test it with 4.1.9.
/Johan
Martijn Tonies wrote:
Hi,
why not try:
SELECT COUNT(s.Id)+COUNT(se.Id)
FROM subs
> Hi,
> why not try:
> SELECT COUNT(s.Id)+COUNT(se.Id)
> FROM subs s
> INNER JOIN subs_erased se ON s.Id=se.Id
> WHERE s.Id=1;
>
> /Johan
This won't return the same result if there's no entries
in "subs_erased" for the ID = 1.
With regards,
Martijn Tonies
Database Workbench - developer tool fo
Hi,
why not try:
SELECT COUNT(s.Id)+COUNT(se.Id)
FROM subs s
INNER JOIN subs_erased se ON s.Id=se.Id
WHERE s.Id=1;
/Johan
Joppe A wrote:
Hello,
I have a small problem that is probably easy to fix but it is to advanced for me as a
"newbe". In my sql-query (below) have I the ID specified twice, I wo
Hello,
I have a small problem that is probably easy to fix but it is to advanced for
me as a "newbe". In my sql-query (below) have I the ID specified twice, I
wonder if there is a easy way to solve it so I only need to write my ID once in
the query?
SELECT (SELECT COUNT(*) FROM subs WHERE id=1
