MySQL 8.0

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create table users_subscriptions (
user_id int, 
subscription_code_id int,
created_at timestamp, 
updated_at timestamp,
expired_at timestamp);

Insert into users_subscriptions values
(1,	69440,"2021-01-23 20:55:37","2021-01-23 20:55:37","2019-01-23 20:55:37"),
(1,	69441,"2021-01-23 21:01:14","2021-01-23 21:01:14","2023-01-23 20:55:37");
SELECT user_id, MAX(expired_at) ed
FROM users_subscriptions
GROUP BY user_id;


user_id
ed


1
2023-01-23 20:55:37


SELECT user_id, MAX(expired_at) ed
FROM users_subscriptions
GROUP BY user_id
HAVING ed > NOW();

SELECT user_id, MAX(expired_at)
FROM users_subscriptions
GROUP BY user_id
HAVING MAX(expired_at) > NOW();


user_id
ed


1
2023-01-23 20:55:37




user_id
MAX(expired_at)


1
2023-01-23 20:55:37


SELECT user_id, MAX(expired_at) ed
FROM users_subscriptions
GROUP BY user_id
HAVING ed < NOW();