从Microsoft SQL Server 2008的列中的每个不同字段中获取前2行

Question

我有一个名为tblHumanResources的表，在该表中我要获取所有行的集合，该行仅包含来自effectiveDate列中每个不同字段的2行（排序方式：升序）：

tblHumanResources表

| empID | effectiveDate |  Company | Description
| 0-123 |    2014-01-23 | DFD Comp | Analyst
| 0-234 |    2014-01-23 | ABC Comp | Manager
| 0-222 |    2012-02-19 | CDC Comp | Janitor
| 0-213 |    2012-03-13 | CBB Comp | Teller
| 0-223 |    2012-01-23 | CBB Comp | Teller

等等。

任何帮助将非常感激。

Answer 1

尝试使用ROW_NUMBER（）函数获取每个组N行：

SELECT * 
FROM
  (
     SELECT t.*,
            ROW_NUMBER() OVER (PARTITION BY effectiveDate 
                               ORDER BY empID) 
            as RowNum
     FROM tblHumanResources as t

  ) as t1
WHERE t1.RowNum<=2
ORDER BY effectiveDate

SQLFiddle演示

没有ROW_NUMBER（）函数的版本假定EmpId在一天中是唯一的：

SELECT * 
FROM tblHumanResources as t
WHERE t.EmpID IN (SELECT TOP 2
                         EmpID 
                    FROM tblHumanResources as t2
                   WHERE t2.effectiveDate=t.effectiveDate
                   ORDER BY EmpID)
ORDER BY effectiveDate

SQLFiddle演示

Answer 2

SELECT TOP 2
        *
FROM    ( SELECT    * ,
                    ROW_NUMBER() OVER ( PARTITION BY effectiveDate ORDER BY effectiveDate ASC ) AS row_num
          FROM      tblHumanResources
        ) AS rows
WHERE   row_num = 1

Answer 3

从tblHumanResources中选择前2个*

Answer 4

对于表中的每个记录，您都选择了有效日期列中与主选择中的当前记录具有相同值的前2行，并且仅当它的empId位于子查询的所选行中时才获取该记录。

select * from tblHumanResources tt
where tt.empID in (select top 2 tt2.empID from tblHumanResources tt2 
                   where tt2.effectiveDate= tt.effectiveDate)

Answer 5

SELECT  TOP 2 *
FROM    (
         SELECT *, ROW_NUMBER() OVER (PARTITION BY effectiveDate ORDER BY effectiveDate ASC) AS row_number
         FROM   tblHumanResources
        ) AS rows
WHERE   row_number = 1

Answer 6

;WITH CTE AS (
SELECT *, ROW_NUMBER() OVER(PARTITION BY effectiveDate ORDER BY effectiveDate ASC) AS RN
FROM tblHumanResources)
SELECT *
FROM CTE
WHERE RN <= 2