我现在有以下声明:
SELECT hl.hour, hl.hourlistex, HOUR(ts.`Timestamp`) as Hour, COUNT(DISTINCT ts.`ForsNr`) as Count, SUM(ts.`TotalAmount`) as SumByReceipt
FROM hourlist hl
LEFT JOIN transactions ts on hl.hour = HOUR(ts.`Timestamp`) AND ts.`SoldDate` = '2016-07-12'
GROUP by hl.hour结果:
hour hourlistex Hour Count SumByReceipt
9 9-10 NULL 0 NULL
10 10-11 NULL 0 NULL
11 11-12 NULL 0 NULL
12 12-13 12 2 152.0000
13 13-14 NULL 0 NULL
14 14-15 14 7 545.0000
15 15-16 15 8 843.0000
16 16-17 16 9 492.0000
17 17-18 17 12 868.0000
18 18-19 18 5 448.0000
19 19-20 NULL 0 NULL
20 20-21 NULL 0 NULL
21 21-22 NULL 0 NULL
22 22-23 NULL 0 NULL我正在寻找一种方法来获得结果,没有开始和结束空值,但包括空值。
像这样:
hour hourlistex Hour Count SumByReceipt
12 12-13 12 2 152.0000
13 13-14 NULL 0 NULL
14 14-15 14 7 545.0000
15 15-16 15 8 843.0000
16 16-17 16 9 492.0000
17 17-18 17 12 868.0000
18 18-19 18 5 448.0000这可以通过一条Mysql语句来实现吗?我熟悉两次辩论之间的争论,但不知道它在我的情况下是否有效。
https://stackoverflow.com/questions/38460151
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