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 the set of dates as part of your query? Something along the lines of:

SELECT COUNT(*) AS Calls, ...
    FROM incidentsm1 RIGHT OUTER JOIN
         (SELECT date_values
            FROM TABLE(('27 Feb 2009'), ('28 Feb 2009'), ('1 Mar 2009'),
                       ('2 Mar 2009'), ('3 Mar 2009'), ('4 Mar 2009'),
                       ('5 Mar 2009')) AS date_list
         )
         ON ...

This is inspired by a sort of hybrid of Informix and DB2 notations and is pretty much guaranteed to be syntactically incorrect in both. Basically, is there a way in your DBMS of creating a literal table on the fly. One possibility - ugly but barely doable - would be to do a 7-way UNION of date literals selected from 'dual' or some table expression that guarantees one row (in Informix terms, SELECT MDY(2,28,2009) FROM "informix".systables WHERE tabid = 1 UNION ...).

Answer 2

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..

Answer 3

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

Answer 4

I would suggest the usage of a date table. With an existing date table in place, you can perform a RIGHT OUTER JOIN to the date table to bring in your missing days.