Sql server 2008 计算差额和百分比

我有以下数据，即某一特定职位的招聘/申请人数，并想找出上一年和本年之间的差异及其增加或减少的百分比

提到了预期输出中括号内使用的公式

Office      year        recruited       applied
Pune        2015            10            15
Pune        2016            7             20
Mumbai      2015            10            23
Mumbai      2016            15            18

我的预期输出应该是：

Office      Difference      %recruited
Pune        -3  (7-10)      -30%(7-10/10)
Mumbai      5(15-10)         50%

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请提供帮助。

如果您使用的是SQL Server 2012或更高版本，则可以使用滞后（或超前）功能。既然你不是，你可以通过CTE获得创造力。这一方法来自于

希望这有帮助

SELECT 'Pune' AS Office,
'2015' AS year,
10 AS recruited,
15 AS applied
INTO #Temp
UNION 
SELECT 'Pune' AS Office,
'2016' AS year,
7 AS recruited,
20 AS applied
UNION 
SELECT 'Mumbai' AS Office,
'2015' AS year,
10 AS recruited,
23 AS applied
UNION 
SELECT 'Mumbai' AS Office,
'2016' AS year,
15 AS recruited,
18 AS applied;

WITH cte AS (
SELECT rownum = ROW_NUMBER() OVER (PARTITION BY t.office ORDER BY t.year), * FROM #Temp t)
SELECT cte.office, cte.recruited - prv.recruited AS DifferenceRecruited,
((cte.recruited - prv.recruited) / CONVERT(FLOAT, prv.recruited) * 100) AS RecruitedChangePercentage,
cte.applied - prv.applied AS DifferenceApplied,
((cte.applied - prv.applied) / CONVERT(FLOAT, prv.applied) * 100) AS AppliedChangePercentage
  FROM cte
LEFT JOIN cte prv ON prv.Office = cte.office AND prv.rownum = cte.rownum - 1
WHERE prv.recruited IS NOT null
ORDER BY cte.Office DESC