如何獲得第一個頂部6個記錄無差異列T-sql？

Question

我得到了一個情況來顯示第一個前6個記錄。前FirstCol前3條記錄和SecondCol後3位。我的查詢是這樣的：

select top 6 [EmpName] 
from [Emp ] 
order by [Salary] Desc 

結果：

[EmpName] 
---------------------- 
Sam   
Pam   
Oliver   
Jam 
Kim 
Nixon 

但我想要的結果看起來像這樣：

FirstCol  SecondCol 
Sam   Jam 
Pam   Kim 
Oliver  Nixon 

Answer 1

您可以使用多個窗口函數做到這一點，這是一種醜陋的，但它會得到你想要的結果：

;with data as 
(
    -- get your Top 6 
    select top 6 empname, salary 
    from emp 
    order by salary desc 
), 
buckets as 
(
    -- use NTILE to split the six rows into 2 buckets 
    select empname, 
    nt = ntile(2) over(order by salary desc), 
    salary 
    from data 
) 
select 
    FirstCol = max(case when nt = 1 then empname end), 
    SecondCol = max(case when nt = 2 then empname end) 
from 
(
    -- create a row number for each item in the buckets to return multiple rows 
    select empname, 
    nt, 
    rn = row_number() over(partition by nt order by salary desc) 
    from buckets 
) d 
group by rn; 

請參閱SQL Fiddle with Demo。這使用函數NTILE，這將採用六行數據集並將其分成兩個桶 - 桶1中的3行和桶2中的3行。NTILE中的(2)用於確定桶的數量。

接下來，我使用row_number()爲每個存儲桶中的每行創建唯一值，這允許您爲每列返回多行。

Answer 2

; WITH TOP_3 AS 
(
    select TOP 3 [EmpName] 
     ,ROW_NUMBER() OVER (ORDER BY [Salary] Desc) rn 
    from [Emp ] 
    order by [Salary] Desc 
), 
    Other3 AS 
(
    SELECT [EmpName] 
      ,ROW_NUMBER() OVER (ORDER BY [Salary] Desc) rn 
    FROM Employees 
    ORDER BY [Salary] DESC OFFSET 3 ROWS FETCH NEXT 3 ROWS ONLY 
) 
SELECT T3.[EmpName] , O3.[EmpName] 
FROM TOP_3 T3 INNER JOIN Other3 O3 
ON T3.RN = O3.RN 
ORDER BY T3.RN ASC