Writing SQL with timestamps - mysql

Question

The data
CREATE TABLE IF NOT EXISTS `transactions` (
`transactions_ts` timestamp ,
`user_id` int(6) unsigned NOT NULL,
`transaction_id` bigint,
`item` varchar(200), PRIMARY KEY(`transaction_id`)
) DEFAULT CHARSET=utf8;
INSERT INTO `transactions` (`transactions_ts`, `user_id`, `transaction_id`,`item` ) VALUES
('2016-06-18 13:46:51.0', 13811335,1322361417, 'glove'),
('2016-06-18 17:29:25.0', 13811335,3729362318, 'hat'),
('2016-06-18 23::07:12.0', 13811335,1322363995,'vase' ),
('2016-06-19 07:14:56.0',13811335,7482365143, 'cup'),
('2016-06-19 21:59:40.0',13811335,1322369619,'mirror' ),
('2016-06-17 12:39:46.0',3378024101,9322351612, 'dress'),
('2016-06-17 20:22:17.0',3378024101,9322353031,'vase' ),
('2016-06-20 11:29:02.0',3378024101,6928364072,'tie'),
('2016-06-20 18:59:48.0',13811335,1322375547, 'mirror');
The question: for each user, show the first item that they ordered (first by time). I assume time as a whole timestamp (not time and date separately).
My attempt
select
min(transactions_ts) as first_trans,
user_id, item
from transactions
group by user_id
order by first_trans;
I am sorry that may be it is a simple question, but one person tells me that my query is entirely wrong. And I have got no other means to test this claim of his
demo fiddle

Answer 1

This is a little bit more complicated than you thought.
To start with: "for each user" would translate to GROUP BY user_id, not to GROUP BY user_id, item.
But with GROUP BY user_id, you'd need an aggregation function saying "the item for the minimum transactions_ts". MySQL doesn't feature such an aggregation function.
The obvious solution is to make this two steps:
Find the first transaction per user
Show the items for these transactions
The query:
select *
from transactions
where (user_id, transactions_ts) in
(
select user_id, min(transactions_ts)
from transactions
group by user_id
);
Another way to word the task is: "Give me the transactions for which no older transaction for the same user exists".
The query:
select *
from transactions t
where not exists
(
select *
from transactions t2
where t2.user_id = t.user_id
and t2.transactions_ts < t.transactions_ts
);

Answer 2

If you are using MySQL 8.0, window function ROW_NUMBER() can be used to adress your use case, as follows:
SELECT transactions_ts, user_id, item
FROM (
SELECT
transactions_ts,
user_id,
item,
ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY transactions_ts) rn
FROM transactions
) x WHERE rn = 1
The inner query ranks each record by ascending timestamp, within groups of records having the same user_id. The outer query filters in the first transaction of each customer.
Demo on DB Fiddle:
transactions_ts | user_id | item
:------------------ | ---------: | :----
2016-06-18 13:46:51 | 13811335 | glove
2016-06-17 12:39:46 | 3378024101 | dress

Answer 3

You can do it using a subquery to get the first transaction_ts for each user:
select user_id, item, transactions_ts
from transactions a
where transactions_ts=(select min(transactions_ts)
from transactions b
where b.user_id=a.user_id)
So your get:
In the inner query get the first transaction time for each user
In the outer query you get the row that has the time you got at point 1