DENSE_RANK according to particular order

Question

Hi I have a table of data I want to output the dense_rank of the names starting from the first group of names according to sorted dates order. e.g.

DROP TABL         


        

Answer 1

You can use:

SELECT DENSE_RANK() OVER (ORDER BY minGrpDate),
       [date], name
FROM (
  SELECT MIN([date]) OVER (PARTITION BY grp) AS minGrpDate,
         [date], name
  FROM (       
    SELECT [date], name,
           ROW_NUMBER() OVER (ORDER BY [date])
           -
           ROW_NUMBER() OVER (PARTITION BY name ORDER BY [date]) AS grp
    FROM mytable) AS t ) AS s
ORDER BY Date

Explanation:

• grp field identifies islands of consecutive records having the same name.
• minGrpDate, which is calculated using grp, is the minimum date of each island.
• Using minGrpDate we can now apply DENSE_RANK() to get required rank.

Note1: The above query handles discontinuities in name field, i.e. the case of non-consecutive fields having the same name.

Note2: The query does not handle the case of different name values sharing the same date value.

Demo here

Answer 2

First rank distinct names ordered by date and then join on the table:

;WITH cte AS(SELECT name, ROW_NUMBER() OVER(ORDER BY MIN(date)) rn 
             FROM dbo.MyTable 
             GROUP BY name)
SELECT c.rn, m.date, m.name
FROM cte c
JOIN dbo.MyTable m ON m.name = c.name
ORDER BY m.date