Find two most recent comparable dates (same weekday and day of month) - sql-server

Question

I've tried searching without much luck. What I'm looking to do is based on any input date, get two most recent dates that had the same day of the month and same weekday as the input date.
For example, date of 02/07/2018 (dd/mm/yyyy) should return 02/04/2018 and 02/10/2017 as they were both Monday the 2nd and were two most recent occurrences.
I have a date table I can work with, I was considering doing a join on itself to get this but I'd appreciate some help.
EDIT: Forgot to mention I'm using SQL Server 2012

Answer 1

Your date table needs to have calculated the day of the week (the number) for each day. Then it's just a simple matter of JOINing, sorting and picking the top 2 rows.
SQL Fiddle
MS SQL Server 2017 Schema Setup:
SETUP THE DATE TABLE
/********************************CALENDAR********************************/
/*
As always, thank you to Aaron Bertrand for the Calendar Table example.
https://www.mssqltips.com/sqlservertip/4054/creating-a-date-dimension-or-calendar-table-in-sql-server/
*/
SET DATEFIRST 7 ; /* SUNDAY */ /* Make sure the week starts on the same day. */
CREATE TABLE datedim (
theDate date PRIMARY KEY
, theDay AS DATEPART(day, theDate) --int
, theWeek AS DATEPART(week, theDate) --int
, theMonth AS DATEPART(month, theDate) --int
, theYear AS DATEPART(year, theDate) --int
, theDOW AS DATEPART(weekday, theDate) --int
, yyyymmdd AS CONVERT(char(8), theDate, 112) /* yyyymmdd */
) ;
/************************************************************************/
/*
Use the catalog views to generate as many rows as we need. This example
creates a date dimension for 1 Sept 2017 to 1 Aug 2018.
*/
INSERT INTO datedim ( theDate )
SELECT d
FROM (
SELECT d = DATEADD(day, s1.rn - 1, '2017-01-01')
FROM
(
SELECT TOP ( DATEDIFF(day, '2017-01-01', '2018-12-31') )
rn = ROW_NUMBER() OVER (ORDER BY sao.object_id)
FROM sys.all_objects sao
) s1
) s2
/************************************************************************/
SETUP TEST DATA
/* TEST DATA */
CREATE TABLE t1 (id int identity, testDate date) ;
INSERT INTO t1 ( testDate )
VALUES
( '2018-04-02' ) /* This */
, ( '2017-10-02' ) /* This */
, ( '2018-04-02' ) /* Duplicate */
, ( '2017-09-27' )
, ( '2018-07-01' )
, ( '2018-05-02' ) /* Same day, Diff DOW */
, ( '2017-09-02' ) /* Same day, Diff DOW */
, ( '2017-10-09' ) /* Diff day, Same DOW */
, ( '2017-01-02' ) /* Same day, Same DOW, Earlier */
, ( null )
;
I added a couple of cases that will get picked up in the early filters but get filtered out by the end. See the notes in the setup.
Main Query:
DECLARE #enteredDate date = '2018-07-02' /* This is the date you entered. */
SELECT TOP 2 s1.testDate
FROM (
SELECT t1.testDate
, ROW_NUMBER() OVER ( PARTITION BY t1.testDate ORDER BY t1.testDate DESC, t1.id DESC ) AS rn
FROM t1
INNER JOIN datedim dd ON t1.testDate = dd.theDate
AND dd.theDay = DATEPART(day,#enteredDate)
AND dd.theDOW = DATEPART(weekday,#enteredDate)
) s1
WHERE rn = 1 /* Get 1st result for a date. */
ORDER BY s1.testDate DESC
In the main query, the first thing we do is use the ROW_NUMBER() window function to make sure we only get one record if there are duplicate date entries. If your data is guaranteed to not have dupe dates, you can skip this step.
[Results][2]:
| testDate |
|------------|
| 2018-04-02 |
| 2017-10-02 |
I think this problem is an excellent example of how useful a Calendar Table can be.

Answer 2

If you have a dates table, then you can use it. The following pseudo-code based on ANSI SQL functions, because date functions depend on the database:
select d.*
from dates d
where extract(dow from d.date) = extract(dow from date '2018-07-02') and
extract(day from d.date) = extract(day from '2018-07-02') and
d.date < date '2018-07-02'
order by d.date desc
fetch first 2 rows only;