Struggling with a simple SQL JOIN - mysql

Question

I'm pretty bad at SQL and have been having some troubles doing somewhat of a UNIQUE join of two tables. The SQL structure is somewhat abysmal, but I didn't design it.
I have two tables:
users
uid, ufn, uln, ue
Where users id = uid.
and
transactions
uid, unit, address, start_date
Basically in the transactions table, there are multiple entries per uid. What I am looking to do is select users.ufn, users.uln, users.ue, transactions.unit, transactions.address based on ONLY the newest start_date. Meaning I will only get ONE result per uid. Currently I'm getting returns for ALL uid entries in the transactions table.
I've tried doing some JOINS, LEFT JOINS, and things with MAX, but have been largely unsuccessful.
SELECT * FROM users JOIN ( SELECT unit, address, start_date FROM transactions GROUP BY uid) as a ON users.tenant_id = a.tenant_id
Is what I tried among a mix of other things.
Any hint as to the right direction would be much appreciated. Thank you!

Answer 1

This will get you close. The problem will be if 2 transactions have the same start date for the same user. But if you don't have that case this should work fine.
select u.ufn,
u.uln,
u.ue,
t.unit,
t.address
from users u
inner join (
select uid,
max(start_date) as newest_start_date
from transactions
group by uid) x
on u.uid = x.uid
inner join transactions t
on t.start_Date = x.newest_start_date
and t.uid = u.uid

Answer 2

Your example SQL has "tenant_id" but that is not in your tables example?
Are you running this once or 10000 times a day?
Try this:
SELECT users.ufn, users.uln, users.ue, transactions.unit, transactions.address
FROM users join transactions on users.uid = transactions.uid
WHERE transactions.UID, transactions.start_date IN
(SELECT UID, MAX(start_date) FROM TRANSACTIONS GROUP BY UID);

Answer 3

Another option is to use an ANTI JOIN on an inequality
select users.ufn,
users.uln,
users.ue,
t.unit,
t.address
from users
INNER JOIN transactions t
ON t.uid = u.uid
LEFT JOIN transactions t1
ON t.uid = t1.uid
and t.start_date < t1.start_date
WHERE
t1.uid is null
Because of t.start_date < t1.start_date and t1.uid is null only records that don't have another record with a greater start_date will be selected
As with MAX() if two or more transaction have start_dates that tie for a user you will get both

Answer 4

This query could work:
SELECT
u.ufn,
u.uln,
u.ue,
t2.unit,
t2.address
FROM
users AS u
INNER JOIN
(
SELECT
uid
, MAX(start_date) AS start_date
FROM
transactions
WHERE
uid = users.uid
) AS t1
INNER JOIN
transaction AS t2 ON t2.uid = t1.uid AND t2.start_date = t1.start_date
Temporary tables are also an option (may be faster, you have to try):
CREATE TEMPORARY TABLE
last_transactions
AS SELECT
uid
, MAX(start_date) AS start_date
FROM
transactions
GROUP BY
uid
;
SELECT
u.ufn,
u.uln,
u.ue,
t2.unit,
t2.address
FROM
users AS u
INNER JOIN
last_transactions AS t1 ON t1.uid = u.uid
INNER JOIN
transaction AS t2 ON t2.uid = t1.uid AND t2.start_date = t1.start_date
P.S.: You should definitely consider adding a primary key to the transactions table. This would allow for a better join clause between t1 and t2. Also, it would prevent duplicates that may occur when multiple start_date occur for the same user.
P.P.S.: wouldn't adding a last_transaction_start_date column to the user table be wiser?

Answer 5

I am not sure if this would be exactly same syntax in MySQL (most likely it will be), but here is how you would do it in SQL server.
Use a rank() function to determine the latest date.
SELECT x.* , y.*
FROM users as x
JOIN
(SELECT *, RANK() Over (Partition By UID Order By Start_Date DESC) as Rank_ FROM Transactions) as y
ON x.uid = y.uid and y.rank_ = 1
Hope this helps.
Cheers!