MySQL AVG(TIMESTAMPDIFF) with GROUP BY

问题

I have two tables user (one) and transaction (many) and I need to get the average time in days from when a user was created to when they made their first transaction. I'm using AVG(TIMESTAMPDIFF) which is working well, except that the GROUP BY returns an average against every user instead of one single average for all unique users in the transaction table. If I remove the GROUP BY, I get a single average figure but it takes into account multiple transactions from users, whereas I just want to have one per user (the first they made).

Here's my SQL:

SELECT AVG(TIMESTAMPDIFF(DAY, u.date_created, t.transaction_date)) AS average
FROM transaction t
LEFT JOIN user u ON u.id = t.user_id
WHERE t.user_id IS NOT NULL AND t.status = 1
GROUP BY t.user_id;

I'd appreciate it if someone can help me return the average for unique users only. It's fine to break the query down into two, but the tables are large so returning lots of data and putting it back in is a no-go. Thanks in advance.

回答1:

SELECT AVG(TIMESTAMPDIFF(DAY, S.date_created, S.transaction_date)) AS average 
FROM (
  SELECT u.date_created, t.transaction_date 
  FROM transaction t 
  INNER JOIN user u ON u.id = t.user_id 
  WHERE t.status = 1 
  GROUP BY t.user_id
  HAVING u.date_created = MIN(u.date_created)
) s

I replaced the LEFT JOIN with an INNER JOIN because I think that's what you want, but it's not 100% equivalant to your WHERE t.user_id IS NOT NULL.
Feel free to put the LEFT JOIN back if need be.

回答2:

select avg( TIMESTAMPDIFF(DAY, u.date_created, min_tdate) ) as average
from user u
inner join 
(select t.user_id, min(t.transaction_date) as min_tdate
 from transaction t
 where t.status=1;
 group by t.user_id
) as min_t
on u.id=min_t.user_id;

来源：https://stackoverflow.com/questions/7269949/mysql-avgtimestampdiff-with-group-by

标签

mysql

group-by

average