Question

我通过SQL Query从SQL Table CONFIRMATION构建每小时报告（最后8小时的生产计数）。查询绝对运行正常并给出正确的结果如下：

SELECT 

(DATENAME(hour,  C.DT_CONFIRMED) + ' - ' + DATENAME(hour, DATEADD(hour, 1, C.DT_CONFIRMED))) as PERIOD,

SUM(C.QT_CONFIRMED) as QT_CONFIRMED

FROM 
    CONFIRMATION C    

WHERE C.DT_CONFIRMED >= DATEADD(hh, -8 , '2015-12-03T11:00:00') 

GROUP BY (DATENAME(hour, C.DT_CONFIRMED) + ' - ' +
          DATENAME(hour, DATEADD(hour, 1, C.DT_CONFIRMED)))

ORDER BY PERIOD

我得到以下结果：

Period      QT_CONFIRMED
8 - 9           4
10 - 11         8

但不是那样，我想要结果跟随形式：

Period   QT_CONFIRMED
2 - 3      0
3 - 4      0
4 - 5      0
5 - 6      0
7 - 8      0
8 - 9      4
9- 10      0
10 - 11    8

基本上，对于QT_CONFIRMED为零的所有时间，我想在报告中显示。

我怎样才能做到这一点？

CONFIRMATION表如下所示：

DT_CONFIRMED                   QT_CONFIRMED      ID_CONFIRMATION
2015-12-03T10:40:43                 5                   1
2015-12-03T10:48:33                 3                   2
2015-12-03T11:03:03                 12                  3

由于

Answer 1

以下内容应该有效。

WITH Periods AS (
    SELECT 8 AS num, (DATENAME(hour,  DATEADD(hour, -9, getDate())) + ' - ' + DATENAME(hour, DATEADD(hour, -8, getDate()))) as PERIOD
    UNION ALL
    SELECT num - 1,  (DATENAME(hour,  DATEADD(hour, -num, getDate())) + ' - ' + DATENAME(hour, DATEADD(hour, -num + 1, getDate()))) as PERIOD
     FROM Periods WHERE num> 0
) ,
 Confrim as (SELECT 
(DATENAME(hour,  C.DT_CONFIRMED) + ' - ' + DATENAME(hour, DATEADD(hour, 1, C.DT_CONFIRMED))) as PERIOD,
SUM(C.QT_CONFIRMED) as QT_CONFIRMED
FROM 
  CONFIRMATION C
WHERE C.DT_CONFIRMED >= DATEADD(hh, -8 , '2015-12-03T11:00:00') 
GROUP BY (DATENAME(hour, C.DT_CONFIRMED) + ' - ' +
          DATENAME(hour, DATEADD(hour, 1, C.DT_CONFIRMED))) )
select P.PERIOD, isnull(C.QT_CONFIRMED, 0) from  Periods P
left join  Confrim C
  on P.PERIOD = C.PERIOD
ORDER BY PERIOD

它只是生成句点，之后会将连接句点留给查询。你可以在这里找到例子： SQL Fiddle sample

此外，您可以将订单更改为 ORDER BY num desc 。之后，您的期间将被正确分类。

Answer 2

您可以使用递归CTE：

https://technet.microsoft.com/en-us/library/ms186243(v=sql.105).aspx

https://dba.stackexchange.com/questions/14661/do-you-know-an-easy-way-to-generate-one-record-for-each-hour-of-the-past-12-hour

SQL中的每小时报告

2 个答案: