how to write the sql statement? - mysql

Question

I have 3 tables as follows in my database. this is used to a application just like foursqure. i need help with the problem of writing the sql statement i have asked in the bottom of this.
thank you
user_details
user_id | fname
----------------
1 | Losh
8 | Dush
9 | Rosh
10 | NELL
friends
user_idf |user_idff
----------------
1 | 8
8 | 9
10 | 1
Check_in
check_in_id |user_id | place | date
--------------------------------------------
1 | 8 | Hotel | 01/01/2012
2 | 9 | Home | 05/01/2012
3 | 1 | Junction | 08/01/2012
4 | 1 | Rest | 11/01/2012
5 | 9 | Hotel | 15/01/2012
6 | 8 | Home | 15/01/2012
i get the user's who are friends with 8 and user 8 details AND the check in places as follows
SELECT a.`user_id`, a.`fname` , b.*
FROM `user_details` a, `check_in` b
WHERE (b.user_id = 8
OR b.user_id in (select user_idf from friends where user_idff = '8' union select user_idff from friends where user_idf = '8')) AND b.user_id = a.user_id
how do i write the sql to select who are friends with 8 and user 8 details AND the last check in place of those users
explanation::
i seeks for a answer such as
user id name place date
1 LOSH Rest 11/01/2012
8 DUSH HOME 15/01/2012
9 ROSH HOTEL 15/01/2012

Answer 1

Join it to the table returned by:
(SELECT `user_id`, `place` FROM Check_in GROUP BY user_id ORDER BY `date` DESC)
That should give you one entry per user, and since it's sorted in reverse by date, that entry should be the most recent.
But when i group by it gives me the first dates not the latest date
How about this:
(SELECT user_id, place
FROM (SELECT * FROM Check_in ORDER BY `date` DESC) tmp
GROUP BY user_id)

Answer 2

SELECT user_id, fname, c.place
FROM user_details u
INNER JOIN (SELECT IF(user_idff = 8, user_idf, user_idff) AS user_id
FROM friends
WHERE (user_idff = 8 OR user_idf = 8)
) f
ON u.user_id = f.user_id
LEFT JOIN (SELECT c1.user_id, place
FROM Check_in c1
LEFT JOIN Check_in c2
ON c1.user_id = c2.user_id AND
c1.date < c2.date
WHERE c2.date IS NULL
) c
ON u.user_id = c.user_id;

Answer 3

This doesn't break ties but it's a straighforward way of answering your question.
EDIT
I just re-read you question and I see that you want user 8 info also. It's not clear whether you want user 8 as a separate row or with info in line with the friends' rows.
select *
from
friends as f inner join check_in as ci on ci.user_id = f.user_idff
inner join user_details as ud on ud.user_id = f.user_idff
inner join user_details as ud8 on ud8.user_id = f.user_idf
where
f.user_idf = 8
and date = (
select max(date)
from friends as f2 inner join check_in as ci on ci.user_id = f2.user_idff
where f2.user_idf = f.user_idf
)
EDIT 2
You request may be a small bit unclear about determining which check-in location to return. Use this option if you want the latest location of each friend individually. The first query finds the most recent location among all friends. Obviously these are two variations on an identical theme.
select *
from
friends as f inner join check_in as ci on ci.user_id = f.user_idff
inner join user_details as ud on ud.user_id = f.user_idff
inner join user_details as ud8 on ud8.user_id = f.user_idf
where
f.user_idf = 8
and date = (
select max(date)
from check_in as ci
where ci.user_id = f.user_idff
)

Answer 4

(SELECT a.user_id, a.place, b.fname, a.date, a.time, a.check_in_id
FROM (SELECT * FROM check_in ORDER BY date DESC) as a, user_details as b
WHERE a.user_id = b.user_id AND (a.user_id in (select user_idf from friends where user_idff = '8' union select user_idff from friends where user_idf = '8') OR a.user_id = 8)
GROUP BY a.user_id)
above query gave me the required answer.
thank you all for the help given