I have a users
table and a payments
table, for each user, those of which have payments, may have multiple associated payments in the payments
table. I would like to select all users who have payments, but only select their latest payment. I'm trying this SQL but i've never tried nested SQL statements before so I want to know what i'm doing wrong. Appreciate the help
SELECT u.*
FROM users AS u
INNER JOIN (
SELECT p.*
FROM payments AS p
ORDER BY date DESC
LIMIT 1
)
ON p.user_id = u.id
WHERE u.package = 1
This question is related to
mysql
sql
select
inner-join
You need to have a subquery to get their latest date per user ID
.
SELECT a.*, c.*
FROM users a
INNER JOIN payments c
ON a.id = c.user_ID
INNER JOIN
(
SELECT user_ID, MAX(date) maxDate
FROM payments
GROUP BY user_ID
) b ON c.user_ID = b.user_ID AND
c.date = b.maxDate
WHERE a.package = 1
This is quite simple do The inner join and then group by user_id and use max aggregate function in payment_id assuming your table being user and payment query can be
select user.id, max(payment.id) from user inner join payment on (user.id = payment.user_id) group by user.id
You can try this:
SELECT u.*, p.*
FROM users AS u LEFT JOIN (
SELECT *, ROW_NUMBER() OVER(PARTITION BY userid ORDER BY [Date] DESC) AS RowNo
FROM payments
) AS p ON u.userid = p.userid AND p.RowNo=1
SELECT u.*, p.*
FROM users AS u
INNER JOIN payments AS p ON p.id = (
SELECT id
FROM payments AS p2
WHERE p2.user_id = u.id
ORDER BY date DESC
LIMIT 1
)
Or
SELECT u.*, p.*
FROM users AS u
INNER JOIN payments AS p ON p.user_id = u.id
WHERE NOT EXISTS (
SELECT 1
FROM payments AS p2
WHERE
p2.user_id = p.user_id AND
(p2.date > p.date OR (p2.date = p.date AND p2.id > p.id))
)
This solutions are better than the accepted answer because they work correctly when there are multiple payments with same user and date. You can try on SQL Fiddle.
There are two problems with your query:
INNER JOIN (SELECT ...) AS p ON ...
. Assuming there are no ties for payments.date
, try:
SELECT u.*, p.*
FROM (
SELECT MAX(p.date) AS date, p.user_id
FROM payments AS p
GROUP BY p.user_id
) AS latestP
INNER JOIN users AS u ON latestP.user_id = u.id
INNER JOIN payments AS p ON p.user_id = u.id AND p.date = latestP.date
WHERE u.package = 1
SELECT u.*
FROM users AS u
INNER JOIN (
SELECT p.*,
@num := if(@id = user_id, @num + 1, 1) as row_number,
@id := user_id as tmp
FROM payments AS p,
(SELECT @num := 0) x,
(SELECT @id := 0) y
ORDER BY p.user_id ASC, date DESC)
ON (p.user_id = u.id) and (p.row_number=1)
WHERE u.package = 1
SELECT U.*, V.* FROM users AS U
INNER JOIN (SELECT *
FROM payments
WHERE id IN (
SELECT MAX(id)
FROM payments
GROUP BY user_id
)) AS V ON U.id = V.user_id
This will get it working
My answer directly inspired from @valex very usefull, if you need several cols in the ORDER BY clause.
SELECT u.*
FROM users AS u
INNER JOIN (
SELECT p.*,
@num := if(@id = user_id, @num + 1, 1) as row_number,
@id := user_id as tmp
FROM (SELECT * FROM payments ORDER BY p.user_id ASC, date DESC) AS p,
(SELECT @num := 0) x,
(SELECT @id := 0) y
)
ON (p.user_id = u.id) and (p.row_number=1)
WHERE u.package = 1
Matei Mihai given a simple and efficient solution but it will not work until put a MAX(date)
in SELECT part so this query will become:
SELECT u.*, p.*, max(date)
FROM payments p
JOIN users u ON u.id=p.user_id AND u.package = 1
GROUP BY u.id
And order by will not make any difference in grouping but it can order the final result provided by group by. I tried it and it worked for me.
SELECT u.*, p.*, max(p.date)
FROM payments p
JOIN users u ON u.id=p.user_id AND u.package = 1
GROUP BY u.id
ORDER BY p.date DESC
Check out this sqlfiddle
@John Woo's answer helped me solve a similar problem. I've improved upon his answer by setting the correct ordering as well. This has worked for me:
SELECT a.*, c.*
FROM users a
INNER JOIN payments c
ON a.id = c.user_ID
INNER JOIN (
SELECT user_ID, MAX(date) as maxDate FROM
(
SELECT user_ID, date
FROM payments
ORDER BY date DESC
) d
GROUP BY user_ID
) b ON c.user_ID = b.user_ID AND
c.date = b.maxDate
WHERE a.package = 1
I'm not sure how efficient this is, though.
Source: Stackoverflow.com