How can I generate a temporary table filled with d

Question

I need to make a temporary table that holds of range of dates, as well as a couple of columns that hold placeholder values (0) for future use. The dates I need are the first day of each month between $startDate and $endDate where these variables can be several years apart.

My original sql statement looked like this:

select dbo.FirstOfMonth(InsertDate) as Month, 0 as Trials, 0 as Sales
into #dates
from customer
group by dbo.FirstOfMonth(InsertDate)

"FirstOfMonth" is a user-defined function I made that pretty much does what it says, returning the first day of the month for the provided date with the time at exactly midnight.

This produced almost exactly what I needed until I discovered there were occasionally gaps in my dates where I had a few months were there were no records insert dates. Since my result must still have the missing months I need a different approach.

I have added the following declarations to the stored procedure anticipating their need for the range of the dates I need ...

declare $startDate set $startDate = select min(InsertDate) from customer
declare $endDate set $endDate = select max(InsertDate) from customer

... but I have no idea what to do from here.

I know this question is similar to this question but, quite frankly, that answer is over my head (I don't often work with SQL and when I do it tends to be on older versions of SQL Server) and there are a few minor differences that are throwing me off.

Answer 1

A starting point of a useful kludge to specify a range or specific list of dates:

SELECT *
FROM 
    (SELECT CONVERT(DateTime,'2017-1-1')+number AS [Date]
     FROM master..spt_values WHERE type='P' AND number<370) AS DatesList
WHERE DatesList.Date IN ('2017-1-1','2017-4-14','2017-4-17','2017-12-25','2017-12-26')

You can get 0 to 2047 out of master..spt_values WHERE type='P', so that's five and a half year's worth of dates if you need it!

Answer 2

Create a table variable containing a date for each month in a year:

declare @months table (reportMonth date, PRIMARY KEY (reportMonth));
declare @start date = '2018', @month int = 0; -- base 0 month
while (@offset < 12)
begin
    insert into @reportMonths select dateAdd(month, @offset, @start);
    select @offset = @offset + 1; 
end

Answer 3

I would probably use a Calendar table. Create a permanent table in your database and fill it with all of the dates. Even if you covered a 100 year range, the table would still only have ~36,525 rows in it.

CREATE TABLE dbo.Calendar (
    calendar_date    DATETIME    NOT NULL,
    is_weekend       BIT         NOT NULL,
    is_holiday       BIT         NOT NULL,
    CONSTRAINT PK_Calendar PRIMARY KEY CLUSTERED (calendar_date)
)

Once the table is created, just populate it once in a loop, so that it's always out there and available to you.

Your query then could be something like this:

SELECT
    C.calendar_date,
    0 AS trials,
    0 AS sales
FROM
    dbo.Calendar C
WHERE
    C.calendar_date BETWEEN @start_date AND @end_date AND
    DAY(C.calendar_date) = 1

You can join in the Customers table however you need to, outer joining on FirstOfMonth(InsertDate) = C.calendar_date if that's what you want.

You can also include a column for day_of_month if you want which would avoid the overhead of calling the DAY() function, but that's fairly trivial, so it probably doesn't matter one way or another.

Answer 4

For SQL Server 2000, this stackoverflow post looks promising for a way to temporarily generate dates calculated off of a start and end date. It's not exactly the same but quite similar. This post has a very in-depth answer on truncating dates, if needed.

In case anyone stumbles on this question and is working in PostgreSQL instead of SQL Server 2000, here is how you might do it there...

PostgreSQL has a nifty series generating function. For your example, you could use this series of all days instead of generating an entire calendar table, and then do groupings and matchups from there.

SELECT current_date + s.a AS dates FROM generate_series(0,14,7) AS s(a);
   dates
------------
 2004-02-05
 2004-02-12
 2004-02-19
(3 rows)

SELECT * FROM generate_series('2008-03-01 00:00'::timestamp,
                              '2008-03-04 12:00', '10 hours');
   generate_series   
---------------------
 2008-03-01 00:00:00
 2008-03-01 10:00:00
 2008-03-01 20:00:00
 2008-03-02 06:00:00
 2008-03-02 16:00:00
 2008-03-03 02:00:00
 2008-03-03 12:00:00
 2008-03-03 22:00:00
 2008-03-04 08:00:00
(9 rows)

I would also look into date_trunc from PostgreSQL using 'month' for the truncator field to maybe refactor your original query to easily match with a date_trunc version of the calendar series.

Answer 5

declare @start datetime
set @start = '2016-09-01'
declare @end datetime
set @end = '2016-09-30'

create table #Date
(
    table_id int identity(1,1) NOT NULL,
    counterDate datetime NULL
);

insert into #Date select top (datediff(D,@start,@end)) NULL from SOME_TABLE
update #Date set counterDate = dateadd(D,table_id - 1, @start)

The code above should populate the table with all the dates between the start and end. You would then just join on this table to get all of the dates needed. If you only needed a certain day of each month, you could dateadd a month instead.

Answer 6

I needed something similar, but all DAYS instead of all MONTHS.

Using the code from MatBailie as a starting point, here's the SQL for creating a permanent table with all dates from 2000-01-01 to 2099-12-31:

CREATE TABLE _Dates (
  d DATE,
  PRIMARY KEY (d)
)
DECLARE @dIncr DATE = '2000-01-01'
DECLARE @dEnd DATE = '2100-01-01'

WHILE ( @dIncr < @dEnd )
BEGIN
  INSERT INTO _Dates (d) VALUES( @dIncr )
  SELECT @dIncr = DATEADD(DAY, 1, @dIncr )
END