如何在单个GROUP BY DAY(date_field)SQL查询中包含空行？

Question

我正在使用MS SQL Server,但欢迎来自其他数据库的比较解决方案.

这是我查询的基本形式.它返回'incidentsm1'表中每天的调用次数:

SELECT 
  COUNT(*) AS "Calls",
  MAX(open_time),
  open_day
FROM 
  (
SELECT
 incident_id,
 opened_by,
 open_time - (9.0/24) AS open_time,
 DATEPART(dd, (open_time-(9.0/24))) AS open_day
   FROM incidentsm1 
   WHERE 
 DATEDIFF(DAY, open_time-(9.0/24), GETDATE())< 7

  ) inc1
GROUP BY open_day

此数据用于绘制条形图,但如果在一周中的某一天没有调用,则没有结果行,因此没有条形图,并且用户就像是,"为什么图表只有六天,从周六到周一跳过？"

不知怎的,我需要UNION ALL每天都有一个空行或类似的东西,但我无法弄明白.

我受限于我可以用一个SQL语句做什么,我只读访问所以我不能创建一个临时表或任何东西.

Answer 1

这样的事怎么样？

SELECT 
  COUNT(incident_id) AS "Calls",
  MAX(open_time),
  days.open_day
FROM
(
  select datepart(dd,dateadd(day,-6,getdate())) as open_day union
  select datepart(dd,dateadd(day,-5,getdate())) as open_day union
  select datepart(dd,dateadd(day,-4,getdate())) as open_day union
  select datepart(dd,dateadd(day,-3,getdate())) as open_day union
  select datepart(dd,dateadd(day,-2,getdate())) as open_day union
  select datepart(dd,dateadd(day,-1,getdate())) as open_day union
  select datepart(dd,dateadd(day, 0,getdate())) as open_day 
) days
left join 
(
 SELECT
   incident_id,
   opened_by,
   open_time - (9.0/24) AS open_time,
   DATEPART(dd, (open_time-(9.0/24))) AS open_day
 FROM incidentsm1 
 WHERE DATEDIFF(DAY, open_time-(9.0/24), GETDATE()) < 7
) inc1 ON days.open_day = incidents.open_day
GROUP BY days.open_day

我只在一个简化的表模式上测试它,但我认为它应该工作.你可能需要修补dateadd的东西..