SQL查询以选择占总数的百分比
我有一个MSSQL表存储,该存储在表中具有以下列:
Storeid, NumEmployees1 125
2 154
3 10
4 698
5 54
6 98
7 87
8 100
9 58
10 897
有人可以帮我进行SQL查询,以产生占雇员总数(NumEmployees)30%的顶级商店(storeID)吗?
回答:
WITH cte
AS (SELECT storeid,
numemployees,
( numemployees * 100 ) / SUM(numemployees) OVER (PARTITION BY 1)
AS
percentofstores
FROM stores)
SELECT *
FROM cte
WHERE percentofstores >= 30
ORDER BY numemployees desc
工作演示
不使用SUM / OVER的替代方法
SELECT s.storeid, s.numemployees FROM (SELECT SUM(numemployees) AS [tots]
FROM stores) AS t,
stores s
WHERE CAST(numemployees AS DECIMAL(15, 5)) / tots >= .3
ORDER BY s.numemployees desc
工作演示
请注意,在第二个版本中,我决定不除以100。这需要强制转换为十进制,否则将隐式转换为int,导致不返回任何记录
同样,我也不太清楚您是否需要这样做,但是您可以将其添加TOP 1
到两个查询中,并将结果限制为商店数量最多的商店,即超过30%的商店
根据您的评论,听起来可以解释凯文(Kevin)
您需要从员工人数最多的商店开始,一直到直到您拥有至少30%的行
这很困难,因为它需要一个运行百分比,并且它有一个装箱问题,但这确实起作用。请注意,我还包括了其他两个测试用例(百分比完全相等,并且刚好在前两个合计的百分比之内)
工作演示
DECLARE @percent DECIMAL (20, 16)SET @percent = 0.3
--Other test values
--SET @percent = 0.6992547128452433
--SET @percent = 0.6992547128452434
;WITH sums
AS (SELECT DISTINCT s.storeid,
s.numemployees,
s.numemployees + Coalesce(SUM(s2.numemployees) OVER (
PARTITION
BY
s.numemployees), 0)
runningsum
FROM stores s
LEFT JOIN stores s2
ON s.numemployees < s2.numemployees),
percents
AS (SELECT storeid,
numemployees,
runningsum,
CAST(runningsum AS DECIMAL(15, 5)) / tots.total
running_percent,
Row_number() OVER (ORDER BY runningsum, storeid ) rn
FROM sums,
(SELECT SUM(numemployees) total
FROM stores) AS tots)
SELECT p.storeID,
p.numemployees,
p.running_percent,
p.running_percent,
p.rn
FROM percents p
CROSS JOIN (SELECT MAX(rn) rn
FROM percents
WHERE running_percent = @percent) exactpercent
LEFT JOIN (SELECT MAX(rn) rn
FROM percents
WHERE running_percent <= @percent) underpercent
ON p.rn <= underpercent.rn
OR ( exactpercent.rn IS NULL
AND p.rn <= underpercent.rn + 1 )
WHERE
underpercent.rn is not null or p.rn = 1
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