Difficult MySQL Query - Getting Max difference between dates - mysql

Question

I have a MySQL table of the following form
account_id | call_date
1 2013-06-07
1 2013-06-09
1 2013-06-21
2 2012-05-01
2 2012-05-02
2 2012-05-06
I want to write a MySQL query that will get the maximum difference (in days) between successive dates in call_date for each account_id. So for the above example, the result of this query would be
account_id | max_diff
1 12
2 4
I'm not sure how to do this. Is this even possible to do in a MySQL query?
I can do datediff(max(call_date),min(call_date)) but this would ignore dates in between the first and last call dates. I need some way of getting the datediff() between each successive call_date for each account_id, then finding the maximum of those.

Answer 1

I'm sure fp's answer will be faster, but just for fun...
SELECT account_id
, MAX(diff) max_diff
FROM
( SELECT x.account_id
, DATEDIFF(MIN(y.call_date),x.call_date) diff
FROM my_table x
JOIN my_table y
ON y.account_id = x.account_id
AND y.call_date > x.call_date
GROUP
BY x.account_id
, x.call_date
) z
GROUP
BY account_id;

Answer 2

CREATE TABLE t
(`account_id` int, `call_date` date)
;
INSERT INTO t
(`account_id`, `call_date`)
VALUES
(1, '2013-06-07'),
(1, '2013-06-09'),
(1, '2013-06-21'),
(2, '2012-05-01'),
(2, '2012-05-02'),
(2, '2012-05-06')
;
select account_id, max(diff) from (
select
account_id,
timestampdiff(day, coalesce(#prev, call_date), call_date) diff,
#prev := call_date
from
t
, (select #prev:=null) v
order by account_id, call_date
) sq
group by account_id
| ACCOUNT_ID | MAX(DIFF) |
|------------|-----------|
| 1 | 12 |
| 2 | 4 |
see it working live in an sqlfiddle

Answer 3

If you have an index on account_id, call_date, then you can do this rather efficiently without variables:
select account_id, max(call_date - prev_call_date) as diff
from (select t.*,
(select t2.call_date
from table t2
where t2.account_id = t.account_id and t2.call_date < t.call_date
order by t2.call_date desc
limit 1
) as prev_call_date
from table t
) t
group by account_id;

Answer 4

Just for educational purposes, doing it with JOIN:
SELECT t1.account_id,
MAX(DATEDIFF(t2.call_date, t1.call_date)) AS max_diff
FROM t t1
LEFT JOIN t t2
ON t2.account_id = t1.account_id
AND t2.call_date > t1.call_date
LEFT JOIN t t3
ON t3.account_id = t1.account_id
AND t3.call_date > t1.call_date
AND t3.call_date < t2.call_date
WHERE t3.account_id IS NULL
GROUP BY t1.account_id
Since you didn't specify, this shows max_diff of NULL for accounts with only 1 call.

Answer 5

SELECT a1.account_id , max(a1.call_date - a2.call_date)
FROM account a2, account a1
WHERE a1.account_id = a2.account_id
AND a1.call_date > a2.call_date
AND NOT EXISTS
(SELECT 1 FROM account a3 WHERE a1.call_date > a3.call_date AND a2.call_date < a3.call_date)
GROUP BY a1.account_id
Which gives :
ACCOUNT_ID MAX(A1.CALL_DATE - A2.CALL_DATE)
1 12
2 4