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Home 2017 October SQL Puzzle: SQL Advance Query – Find date range for same Market Rank

SQL Puzzle: SQL Advance Query – Find date range for same Market Rank

This article is half-done without your Comment! *** Please share your thoughts via Comment ***

Check the below input data and require output data to find the start date and end date range for the same market rank and if market rank change, only generates a new range row.

Input data:

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rankdate rankno
---------- -----------
2017-01-01 24000
2017-02-01 24000
2017-03-01 26000
2017-04-01 26000
2017-05-01 26000
2017-06-01 26000
2017-07-01 29000
2017-08-01 29000
2017-09-01 29000
2017-10-01 29500
2017-11-01 29500
2017-11-01 30000

Expected output:

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StartDate EndDate rankno
---------- ---------- -----------
2017-01-01 2017-02-01 24000
2017-03-01 2017-06-01 26000
2017-07-01 2017-09-01 29000
2017-10-01 2017-11-01 29500
2017-11-01 2017-11-01 30000

Create a table with sample data:

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CREATE TABLE MarketRank
(
rankdate date
,rankno int
)
GO
 
INSERT INTO MarketRank
VALUES
('2017-01-01', 24000)
,('2017-02-01', 24000)
,('2017-03-01', 26000)
,('2017-04-01', 26000)
,('2017-05-01', 26000)
,('2017-06-01', 26000)
,('2017-07-01', 29000)
,('2017-08-01', 29000)
,('2017-09-01', 29000)
,('2017-10-01', 29500)
,('2017-11-01', 29500)
,('2017-11-01', 30000)
GO

Solution 1:

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WITH CTE1 AS
(
SELECT rankdate , rankno , ROW_NUMBER() OVER (ORDER BY rankdate) Rnk
FROM MarketRank
)
,CTE2 AS
(
SELECT *, CASE WHEN rankno = lag(rankno) over(order by rnk) THEN 0 ELSE 1 END rankcols
FROM CTE1 c2
)
,CTE3 AS
(
SELECT *,SUM(rankcols) over(order by rnk) Grouper
FROM CTE2 c2
)
SELECT
MIN(rankdate) StartDate
,MAX(rankdate) EndDate
,MAX(rankno) rankno
FROM CTE3 GROUP BY Grouper

Solution 2:

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WITH CTE1 AS ( SELECT rankdate , rankno , ROW_NUMBER() OVER (ORDER BY rankdate) Rnk FROM MarketRank )
,CTE2 AS
(
SELECT
*,
CASE
WHEN rankno = ( SELECT rankno from CTE1 c3 WHERE c3.rnk = ( SELECT MAX(c1.rnk) FROM CTE1 c1 WHERE c1.rnk < c2.rnk ))
THEN 0 ELSE 1 END Identifier
,SUM(CASE
WHEN rankno = ( SELECT rankno from CTE1 c3 WHERE c3.rnk = ( SELECT MAX(c1.rnk) FROM CTE1 c1 WHERE c1.rnk < c2.rnk ))
THEN 0 ELSE 1 END) OVER (ORDER BY rnk ) rankcols
FROM CTE1 c2
)
SELECT MIN(rankdate) StartDate , MAX(rankdate) EndDate , MAX(rankno) rankno
FROM CTE2 GROUP BY rankcols

Oct 28, 2017Anvesh Patel
SQL Puzzle: SQL Advance Query - Count the total Working days from the given datesSQL Puzzle: SQL Advance Query - Print START before group and Print END after group
Comments: 2
  1. Dinesh.IS
    November 7, 2017 at 5:08 am

    –Method-1
    Select Min(RankDate) As StartDate,Max(RankDate) As EndDate,B.RankNo
    From MarketRank A
    Inner JOin (Select Distinct RankNo
    From MarketRank) B On A.rankno=B.rankno
    Group By B.RankNo

    –Method-2
    Select Min(RankDate) As StartDate,Max(RankDate) As EndDate,RankNo
    From (Select RankDate,RankNo
    From MarketRank ) A
    Group By RankNo

    ReplyCancel
  2. Lakshman
    April 27, 2019 at 4:19 pm

    SELECT MIN(RankDate) AS StartDate,MAX(RankDate) AS EndDate,RankNo
    FROM MarketRank
    GROUP BY RankNo

    ReplyCancel

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Anvesh Patel
Anvesh Patel

Database Engineer

October 28, 2017 2 Comments SQL PuzzleAnvesh Patel, database, database research and development, dbrnd, SQL Advance Query, SQL Interview, SQL Problem, SQL Programming, SQL Puzzle, SQL Query, SQL Tips and Tricks
About Me!

I'm Anvesh Patel, a Database Engineer certified by Oracle and IBM. I'm working as a Database Architect, Database Optimizer, Database Administrator, Database Developer. Providing the best articles and solutions for different problems in the best manner through my blogs is my passion. I have more than six years of experience with various RDBMS products like MSSQL Server, PostgreSQL, MySQL, Greenplum and currently learning and doing research on BIGData and NoSQL technology. -- Hyderabad, India.

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