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

Answer 1

Can you create a table variable with the dates that you need and then RIGHT JOIN onto it? For example,

DECLARE @dateTable TABLE ([date] SMALLDATETIME)

INSERT INTO @dateTable
VALUES('26 FEB 2009')
INSERT INTO @dateTable
VALUES('27 FEB 2009')
-- etc

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
   RIGHT JOIN @dateTable dates
   ON incidentsm1.open_day = dates.date
   WHERE 
 DATEDIFF(DAY, open_time-(9.0/24), GETDATE())< 7

  ) inc1
GROUP BY open_day

The more ideal situation however, would be to have a table object with the dates in