Comments on: SQL Puzzle: SQL Advance Query – Solve the challenge of DISTINCT https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-solve-the-challenge-of-distinct/ BIGData | NoSQL | MSSQL | MySQL | PostgreSQL Thu, 21 Mar 2019 12:47:38 +0000 hourly 1 https://wordpress.org/?v=7.0.6 By: Marcel Hofman https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-solve-the-challenge-of-distinct/#comment-6921 Thu, 21 Mar 2019 12:47:38 +0000 https://www.dbrnd.com/?p=6184#comment-6921 ; WITH firstcycle as
(
SELECT
ROW_NUMBER () OVER ( ORDER BY (SELECT 0) ) AS n
, col3
, col1 AS newcolumntwo
, NULL AS newcolumnone
FROM tbl_DataColumn
ORDER BY (SELECT 0)
OFFSET 0 ROWS
FETCH NEXT (SELECT COUNT(DISTINCT col3) FROM tbl_DataColumn) ROWS ONLY
)
,
secondcycle AS
(
SELECT
ROW_NUMBER () OVER ( ORDER BY (SELECT 0) ) AS n
, col3
, NULL AS newcolumntwo
, col1 AS newcolumnone
FROM tbl_DataColumn
ORDER BY (SELECT 0)
OFFSET (SELECT COUNT(DISTINCT col3) FROM tbl_DataColumn) ROWS
FETCH NEXT (SELECT COUNT(DISTINCT col3) from tbl_DataColumn) ROWS ONLY
)
SELECT s.newcolumnone AS col1, f.newcolumntwo AS col2, f.col3
FROM firstcycle f
LEFT JOIN secondcycle s
ON s.n -(SELECT COUNT(DISTINCT col3) FROM tbl_DataColumn) = f.n

]]> By: Anvesh Patel https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-solve-the-challenge-of-distinct/#comment-5892 Mon, 09 Oct 2017 14:29:26 +0000 https://www.dbrnd.com/?p=6184#comment-5892 Perfect!

]]> By: Anvesh Patel https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-solve-the-challenge-of-distinct/#comment-5891 Mon, 09 Oct 2017 14:29:06 +0000 https://www.dbrnd.com/?p=6184#comment-5891 Good!

]]> By: Dinesh I.S https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-solve-the-challenge-of-distinct/#comment-5890 Mon, 09 Oct 2017 13:29:03 +0000 https://www.dbrnd.com/?p=6184#comment-5890 —While Looping We can get Result

Declare @TableResult Table(Col1 Int,Col2 Int,Col3 Int)
Delete From @TableResult
Declare @Amt As Int
Select @Amt=Min(Col3) From TBL_DataColumn

While @Amt Is Not Null
Begin
Insert Into @TableResult
(Col1,Col2,Col3)
Select *,@Amt From (
Select Row_Number()Over(Order By Col1) As SLNo,Col1
From TBL_DataColumn Where Col3=@Amt)A
PIVOT(Max(Col1) For SLNo In([1],[2])) A
Select @Amt=Min(Col3) From TBL_DataColumn Where Col3>@Amt
End
Select * From @TableResult

]]> By: Dinesh I.S https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-solve-the-challenge-of-distinct/#comment-5889 Mon, 09 Oct 2017 13:17:21 +0000 https://www.dbrnd.com/?p=6184#comment-5889 ;
WITH C1
As
(Select Row_Number()Over(Order By Col3) As SLNo,Col3
From (Select Distinct Col3
From TBL_DataColumn) A) ,
C2
As
(Select Row_Number()Over(PARTITION By GSLNo Order By Col1) As SLNo,GSLNo,Col1
From (Select NTILE(2)Over(Order By Col1) As GSLNo,Col1 From TBL_DataColumn) A )
Select C2.Col1 As Col2,C3.Col1 As Col3,C1.Col3
From C1 Inner Join C2 On C1.SLNo=C2.SLNo
Inner JOin C2 As C3 On C3.SLNo=C1.SLNo
Where C2.GSLNo=1
And C3.GSLNo=2

]]>