mysql - 按天分组和总计工作时间

我在写一个管理工作时间的软件。因为每一天都有不同的活动。
我想把过去7天每个工人的工作时间都提取出来。
示例行

+----+------------+-------+----------+----------+
| id |    date    | order | operator | duration |
+----+------------+-------+----------+----------+
| 37 | 2016-06-12 |    27 |        1 |      180 |
| 38 | 2016-06-12 |    28 |        3 |      390 |
| 39 | 2016-06-12 |    27 |        1 |      480 |
| 40 | 2016-06-04 |    21 |        2 |      120 |
| 41 | 2016-05-07 |    27 |        1 |       90 |
| 42 | 2016-06-07 |    27 |        1 |      150 |
+----+------------+-------+----------+----------+

询问

SELECT SUM(`duration`) as `hours_per_day`
FROM `sheets`
WHERE `operator` = 1 AND `date` = DATE_ADD(CURDATE(), INTERVAL -7 DAY)
ORDER BY `date` DESC

预期结果：

+------------------------+--------+--------+--------+--------+--------+--------+--------+
| Operator: Avareage Joe                                                                |
+------------------------+--------+--------+--------+--------+--------+--------+--------+
| Day:                   | 01 jul | 02 jul | 03 jul | O4 jul | 05 jul | 06 jul | 07 jul |
| Hours:                 | 8      | 7      | 9      | 8      | 9      | 8      | 6      |
+------------------------+--------+--------+--------+--------+--------+--------+--------+

最佳答案

SQL Fiddle Demo

SELECT
 CURDATE(),
 SUM(CASE WHEN CURDATE()                  = `date` THEN `duration` ELSE 0 END) as today    ,
 SUM(CASE WHEN CURDATE() - INTERVAL 1 DAY = `date` THEN `duration` ELSE 0 END) as yesterday,
 SUM(CASE WHEN CURDATE() - INTERVAL 2 DAY = `date` THEN `duration` ELSE 0 END) as `today - 2`,
 SUM(CASE WHEN CURDATE() - INTERVAL 3 DAY = `date` THEN `duration` ELSE 0 END) as `today - 3`,
 SUM(CASE WHEN CURDATE() - INTERVAL 4 DAY = `date` THEN `duration` ELSE 0 END) as `today - 4`,
 SUM(CASE WHEN CURDATE() - INTERVAL 5 DAY = `date` THEN `duration` ELSE 0 END) as `today - 5`,
 SUM(CASE WHEN CURDATE() - INTERVAL 6 DAY = `date` THEN `duration` ELSE 0 END) as `today - 6`
FROM `work`
where operator = 1

输出