Get response time per day of week - mysql

Question

I have 3 tables:
CREATE TABLE `ticket` (
`tid` int(11) NOT NULL AUTO_INCREMENT,
`sid` varchar(50) NOT NULL,
`open_date` datetime NOT NULL,
PRIMARY KEY (`tid`),
KEY `sid` (`sid`,`open_date`),
KEY `open_date` (`open_date`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE `ticket_reply` (
`rid` int(11) NOT NULL AUTO_INCREMENT,
`tid` int(11) NOT NULL,
`reply_date` datetime NOT NULL,
PRIMARY KEY (`rid`),
KEY `tid` (`tid`,`reply_date`),
KEY `reply_date` (`reply_date`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE `subscription` (
`sid` varchar(50) NOT NULL,
`response_time` int(11) NOT NULL DEFAULT '24',
PRIMARY KEY (`sid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
I'm trying to get the sum of response times the first ticket reply is from when the ticket was opened and group it by DAYNAME (maybe by MONTH also). Currently I have this SQL:
SELECT
t.tid,
DAYNAME(t.open_date) AS day_opened,
SUM(TIMESTAMPDIFF(MINUTE, t.open_date, tr.reply_date)) AS num_min,
SUM(s.response_time * 60) AS response_time_min
FROM ticket t
INNER JOIN ticket_reply tr ON tr.tid = t.tid
INNER JOIN subscription s ON s.sid = t.sid
GROUP BY
t.tid #group by tid as ticket_reply may return many
ORDER BY t.open_date DESC;
So first challenge I have is getting the first ticket_reply row which I solved by GROUP BY, I tried to get a subquery in the join but it was still returning a row per ticket_reply row.
So now I want to start grouping by DAYNAME and maybe MONTH but if I add it to the GROUP BY it doesn't group:
GROUP BY
t.tid,
DAYNAME(t.open_date)
Have tried DAYNAME before tid but that didn't make any difference.
So I have a couple questions, is there a better way to get the first row in ticket_reply and then group by the DAYNAME? I have a feeling getting the first row in a subquery may fix the grouping.

Answer 1

It is grouping, but because you have t.tid in the GROUP BY clause, and the tid column is unique in the ticket table, multiple rows from ticket are not getting collapsed, each is going to be on its own row.
It's not exactly clear what result you want to return.
(The SUM(s.response_time) expression in the SELECT list seems a bit odd, given that you could be matching multiple rows from ticket_reply.)
Given your existing statement, it looks like you might want to use an inline view to return the "earliest" reply_date for each ticket, in place of the reference to the ticket_reply table.
JOIN /*ticket_reply*/
( SELECT r.tid
, MIN(r.reply_date) AS reply_date
FROM ticket_reply r
GROUP BY r.tid
) tr ON tr.tid = t.tid
(Unfortunately, materializing the inline view (populating and accessing an intermediate "derived table") can be the source of a performance issue.)
As another option, rather than performing a JOIN operation, you could consider using a correlated subquery in the SELECT list. That is, in place of the reference to tr.reply_date, you could do something like:
(SELECT MIN(r.reply_date) FROM ticket_reply r WHERE r.tid = t.tid)
and remove the JOIN to the ticket_reply table.
But, repeated execution of that subquery (once for each row returned), can also be a performance issue for large sets.
But the "big" question is whether you need to add up s.response_time for each occurrence of a matching ticket_reply (as your current query is doing), or whether you just need to include the s.response_time once for each ticket?
That is, if there are three ticket_reply for a given ticket, do we need to "triple" the value of response_time that we add to the row?
If you need to include the response_time in the total for each `ticket_reply, then this:
SELECT DAYNAME(t.open_date) AS day_opened
, SUM(TIMESTAMPDIFF(MINUTE, t.open_date, tr.reply_date)) AS num_min
, SUM(s.response_time) * 60 * tr.cnt_replies AS response_time_min
FROM ticket t
JOIN ( SELECT r.tid
, MIN(r.reply_date) AS reply_date
, COUNT(1) AS cnt_replies
FROM ticket_reply r
GROUP BY r.tid
) tr
ON tr.tid = t.tid
JOIN subscription s
ON s.sid = t.sid
GROUP
BY day_opened
If you only need to include the response_time in the total once for each ticket, remove the references to cnt_replies:
SELECT DAYNAME(t.open_date) AS day_opened
, SUM(TIMESTAMPDIFF(MINUTE, t.open_date, tr.reply_date)) AS num_min
, SUM(s.response_time) * 60 AS response_time_min
FROM ticket t
JOIN ( SELECT r.tid
, MIN(r.reply_date) AS reply_date
FROM ticket_reply r
GROUP BY r.tid
) tr
ON tr.tid = t.tid
JOIN subscription s
ON s.sid = t.sid
GROUP
BY day_opened
To GROUP BY month, just change the first expression in the SELECT list, and the reference in the GROUP BY clause.