How do I include empty rows in a single GROUP BY DAY(date_field) SQL query?

Question

I'm using MS SQL Server but welcome comparitive solutions from other databases.

This is the basic form of my query. It returns the number of calls per day from the 'incidentsm1' table:

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

This data is used to draw a bar graph, but if there were no calls on a given day of the week, there is no result row and thus no bar, and the user is like, "why does the graph only have six days and skip from Saturday to Monday?"

Somehow I need to UNION ALL with a blank row from each day or something like that, but I can't figure it out.

I am constrained to what I can do with one SQL statement and I have readonly access so I can't create a temporary table or anything.

Answer 1

How about something like this?

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

I've only tested it on a simplified table schema, but I think it should work. You might need to tinker with the dateadd stuff..