Thanks to all that responded about my counting question. Gleb
Paharenko's solution using the sub-select and the LEFT JOINS was the
closest to what I needed.
All the best,
Dean...K...
--
Dean Karres / karres at itg dot uiuc dot edu / www.itg.uiuc.edu
Im
Hello.
What do you think about this:
SELECT u.name,
IFNULL(s.skill_name,'user doesn\'t have any
skill') AS SKILL,
(
SELECT COUNT(skill_id)
I could not help to resist to create the tables and
try the hint, which is pretty close to the working
solution.
I just had to replace "order by user.id" with "group
by user.name skill.skill_name"
regards,
esv.
--- Parag Agrawal <[EMAIL PROTECTED]> wrote:
> Try this out:
>
> select user.nam
Try this out:
select user.name, skill.skill_name, count(user_skill1.skill_id)
from user , skill, user_skill , user_skill as user_skill1
where user.id=user_skill.user_id and user_skill.skill_id=skill.id and
skill.id = user_skill1.skill_id
order by user.id.
I think this would work.
--
Parag
B
Hi,
I have three simple tables:
user
{
id int PRIMARY
name varchar UNIQUE
}
skill
{
id int PRIMARY
skill_name varchar UNIQUE
}
user_skill
{
user_id int
skill_id int
UNIQUE KEY id (user_id, skill_id)
}
Each user can have zero or more associated skills. So you might have
these a
Try this:
select delivery, count(*) as ticketcount
from ticketsales
where delivery="post" or delivery="pickup"
group by delivery
Andy
> -Original Message-
> From: Ville Mattila [mailto:[EMAIL PROTECTED]
> Sent: 03 July 2003 11:28
> To: [EMAIL PROTECT
Hi there,
We are currently having a large festival here in Ikaalinen, Finland
(www.satahamesoi.fi) and we have our booking system based on MySQL.
I should find out how many tickets are sold in each concert with different
delivery methods. How could I make a count like this:
SELECT COUNT(delivery=
