Jasper Bryant-Greene wrote:
Michael Stassen wrote:
Not exactly. They aren't the same. COUNT(id) counts distinct values
of id, while COUNT(*) simply counts rows. [snip]
Actually, COUNT(id) counts non-NULL values of id. COUNT(DISTINCT id)
would count distinct values.
http://dev.mysql.com
Michael Stassen wrote:
Not exactly. They aren't the same. COUNT(id) counts distinct values of
id, while COUNT(*) simply counts rows. [snip]
Actually, COUNT(id) counts non-NULL values of id. COUNT(DISTINCT id)
would count distinct values.
http://dev.mysql.com/doc/mysql/en/group-by-functi
Sebastian wrote:
Gleb Paharenko wrote:
Hello.
What about this:
UPDATE news
SET comments =
( SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Hi. i came up with a similar query last night, but i didnt use group by.
I have question, is it better to use COUNT(*
Nuno Pereira <[EMAIL PROTECTED]> wrote on 08/08/2005 01:49:44 PM:
> Sebastian wrote:
> > Nuno Pereira wrote:
> >
> >> [EMAIL PROTECTED] wrote:
> >>
> >>> Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005
> >>> 10:53:55 AM:
> >>>
> >>>
> Gleb Paharenko wrote:
>
>
> > Hello.
> >
Sebastian wrote:
Nuno Pereira wrote:
[EMAIL PROTECTED] wrote:
Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005
10:53:55 AM:
Gleb Paharenko wrote:
Hello.
What about this:
UPDATE news
SET comments =
(
SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005 01:19:32 PM:
> Nuno Pereira wrote:
>
> > [EMAIL PROTECTED] wrote:
> >
> >> Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005
> >> 10:53:55 AM:
> >>
> >>
> >>> Gleb Paharenko wrote:
> >>>
> >>>
> Hello.
>
>
>
> What about
Nuno Pereira wrote:
[EMAIL PROTECTED] wrote:
Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005
10:53:55 AM:
Gleb Paharenko wrote:
Hello.
What about this:
UPDATE news
SET comments =
(
SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Hi. i came up
[EMAIL PROTECTED] wrote:
Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005 10:53:55 AM:
Gleb Paharenko wrote:
Hello.
What about this:
UPDATE news
SET comments =
(
SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Hi. i came up with a similar q
[EMAIL PROTECTED] wrote:
Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005 10:53:55 AM:
Gleb Paharenko wrote:
Hello.
What about this:
UPDATE news
SET comments =
(
SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Hi. i came up with a similar q
Sebastian <[EMAIL PROTECTED]> wrote on 08/08/2005 10:53:55 AM:
> Gleb Paharenko wrote:
>
> >Hello.
> >
> >
> >
> >What about this:
> >
> >
> >
> >UPDATE news
> >
> >SET comments =
> >
> >(
> >
> > SELECT COUNT(id)
> >
> > FROM comments
> >
> > WHERE newsid = news.id
> >
> > GROUP BY news
Gleb Paharenko wrote:
Hello.
What about this:
UPDATE news
SET comments =
(
SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Hi. i came up with a similar query last night, but i didnt use group by.
I have question, it it better to use COUNT
Hello.
No, it isn't. Here is the test case:
create table comments(id int auto_increment,newsid int,primary key(id));
create table news(id int,comments int);
insert into news(id) values('1'),(2),(3),(4),(5);
insert into comments(newsid) values(1),(1),(2),(4),(4),(4);
update news set comm
--- Gleb Paharenko <[EMAIL PROTECTED]> wrote:
> Hello.
>
> What about this:
>
> UPDATE news
> SET comments =
> (
> SELECT COUNT(id)
> FROM comments
> WHERE newsid = news.id
> GROUP BY newsid
> );
>
>
Isn't the previous query going to update the
news.comments wit
Hello.
What about this:
UPDATE news
SET comments =
(
SELECT COUNT(id)
FROM comments
WHERE newsid = news.id
GROUP BY newsid
);
Sebastian <[EMAIL PROTECTED]> wrote:
> I have two tables: news and comments.
>
> i want to count the number of comments f
I have two tables: news and comments.
i want to count the number of comments for each newsid and update the
count in news.comment
comments.newsid belongs to news.id
can i do this with sub queries? im using v4.1.x
i have many records in the comments table so im not sure what the most
efficien
15 matches
Mail list logo
