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Home/ Questions/Q 6032241
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T05:18:18+00:00 2026-05-23T05:18:18+00:00

I have the following table VendorID EMP1A EMP2A EMP1B EMP2B ———– ———– ———– ———–

  • 0

I have the following table

VendorID    EMP1A       EMP2A       EMP1B       EMP2B
----------- ----------- ----------- ----------- -----------
1           4           3           5           4
2           4           1           5           5
3           4           3           5           4
4           4           2           5           5
5           5           1           5           5

I want to UNPIVOT in pairs so I get the “A” columns on one row and the “B” columns on another

vendorid    employee  orders      employee  orders
----------- --------- ----------- --------- -----------
1           EMP1A     4           EMP2A     3
1           EMP1B     5           EMP2B     4
2           EMP1A     4           EMP2A     1
2           EMP1B     5           EMP2B     5
3           EMP1A     4           EMP2A     3
3           EMP1B     5           EMP2B     4
4           EMP1A     4           EMP2A     2
4           EMP1B     5           EMP2B     5
5           EMP1A     5           EMP2A     1
5           EMP1B     5           EMP2B     5

This works but it seems like I’m working too hard

DECLARE @pvt AS TABLE( 
  VendorID INT, 
  EMP1A    INT, 
  EMP2A    INT, 
  EMP1B    INT, 
  EMP2B    INT); 

INSERT INTO @pvt VALUES (1,4,3,5,4),
(2,4,1,5,5),
(3,4,3,5,4),
(4,4,2,5,5),
(5,5,1,5,5)

;WITH piv1 
     AS (SELECT vendorid, 
                employee, 
                orders 
         FROM   (SELECT vendorid, 
                        EMP1A, 
                        EMP1B 
                 FROM   @pvt) p 
                 UNPIVOT (orders FOR employee IN (EMP1A, EMP1B) ) 
                AS 
                unpvt), 
     piv2 
     AS (SELECT vendorid, 
                employee, 
                orders 
         FROM   (SELECT vendorid, 
                        EMP2A, 
                        EMP2B 
                 FROM   @pvt) p UNPIVOT (orders FOR employee IN (EMP2A, EMP2B) ) 
                AS 
                unpvt) 
SELECT piv1.vendorid, 
       piv1.employee, 
       piv1.orders, 
       piv2.employee, 
       piv2.orders 
FROM   piv1 
       INNER JOIN piv2 
         ON piv1.vendorid = piv2.vendorid 
            AND RIGHT(piv1.employee, 1) = RIGHT(piv2.employee, 1) 
WHERE
    piv1.orders > piv2.orders

Note: This is a simplified example and there’s actually 25 pairs that need to be converted to rows and I also want easily be able to filter

e.g. adding WHERE piv1.orders = piv2.orders produces

vendorid    employee  orders      employee  orders
----------- --------- ----------- --------- -----------
2           EMP1B     5           EMP2B     5
4           EMP1B     5           EMP2B     5
5           EMP1B     5           EMP2B     5
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T05:18:18+00:00Added an answer on May 23, 2026 at 5:18 am

    turns out I was working too hard

    WITH piv1 
         AS (SELECT vendorid, 
                    employee, 
                    orders 
             FROM   @pvt p UNPIVOT (orders FOR employee IN (emp1a, emp1b, 
                                                            emp2a, emp2b) ) 
                    AS unpvt) 
    SELECT piv1.vendorid, 
           piv1.employee, 
           piv1.orders, 
           piv2.employee, 
           piv2.orders 
    FROM   piv1 
           INNER JOIN piv1 piv2 
             ON piv1.vendorid = piv2.vendorid 
                AND piv1.employee <> piv2.employee 
                AND RIGHT(piv1.employee, 1) = RIGHT(piv2.employee, 1) 
    WHERE  Substring(piv1.employee, 4, 1) <> '2' 
    ORDER  BY piv1.vendorid, 
              piv1.employee 
    

    Now adding pairs just requires updating the UNPIVOT
    e.g.

    @pvt p UNPIVOT (orders FOR employee IN (emp1a, emp1b,
                                            emp2a, emp2b,
                                            emp1c, emp2c,
                                            emp1d, emp2d) )
    
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