как рассчитать среднее время интервала
+-------------+----------+----------+--------+------------------+
| customer_id | date | time | answer | missed_call_type |
+-------------+----------+----------+--------+------------------+
| 101 | 2018/8/3 | 12:13:00 | no | employee |
| 102 | 2018/8/3 | 12:15:00 | no | customer |
| 103 | 2018/8/3 | 12:20:00 | no | employee |
| 102 | 2018/8/3 | 15:15:00 | no | customer |
| 101 | 2018/8/3 | 18:15:00 | no | employee |
| 105 | 2018/8/3 | 18:18:00 | no | customer |
| 102 | 2018/8/3 | 19:18:00 | no | employee |
+-------------+----------+----------+--------+------------------+
У меня есть таблица, которая выглядит так и хотела рассчитать среднее время для тех, кто не ответил на телефон. В этом примере среднее время интервала:
{(18: 15: 00-12: 13: 00) + [(19: 18: 00-15: 15: 00) + (15: 15: 00-12: 15: 00)]/2}/2
Ниже приведен код для MSSQL,
CREATE TABLE customer_data (customer_id BIGINT, date DATE, time time, answer VARCHAR(100), missed_call_type VARCHAR(100));
INSERT INTO customer_data
VALUES
(101, '2018/8/3', '12:13:00', 'no', 'employee'),
(102, '2018/8/3', '12:15:00', 'no', 'customer'),
(103, '2018/8/3', '12:20:00', 'no', 'employee'),
(102, '2018/8/3', '15:15:00', 'no', 'customer'),
(101, '2018/8/3', '18:15:00', 'no', 'employee'),
(105, '2018/8/3', '18:18:00', 'no', 'customer'),
(102, '2018/8/3', '19:18:00', 'no', 'employee')
select cd.customer_id, answer, missed_call_type,
CAST(CAST(cd.date as VARCHAR(10))+' ' +CAST(cd.time as VARCHAR(10)) as datetime) as date,
ROW_NUMBER() OVER(PARTITION BY cd.customer_id ORDER BY date desc, time desc) as ranks
INTO #temP
from customer_data cd
order by cd.customer_Id, ranks;
select AVG(DATEDIFF(MINUTE, x1.date, x2.date)) as avg_mins
from #temP x1
INNER JOIN #temP x2 ON x1.customer_id = x2.customer_id
WHERE x2.ranks = (x1.ranks-1)