如何根据列值获得百分比？

Question

更新 2015 年 5 月 18 日

有一个列有客户姓名的列。我怎样才能得到基于每个日期一个客户的百分比？例如

    CustomerName     Date
    Sam              04/29/15
    Joy              04/29/15
    Tom              04/29/15
    Sam              04/29/15
    Oly              04/29/15
    Joy              04/29/15
                     04/29/15
    Sam              04/29/15
                     04/29/15
    Sam              04/29/15
    Oly              04/29/15
    Sam              04/29/15
    Oly              04/30/15
    Joy              05/01/15

请注意，我的列有 12 条记录，其中 2 条为空白，但它们不计算百分比，只计算有名称的记录。我想知道 Sam 在总数中所占的百分比（在本例中为 10 条记录，因此 Sam % 将为 50）。

查询应该return

Date             Percentage
04/29/15            50
04/30/15            0
05/01/15            0

更新

我真的不关心其他客户，所以让我们把他们当作一个。只需要知道 Sam 在总列表中的百分比。

任何帮助将不胜感激。谢谢

Answer 1

你会做这样的事情

SELECT (COUNT(*) * 100) / (SELECT COUNT(*) FROM Customer WHERE CustomerName <> '' AND Date = '04/29/15')
FROM Customer
WHERE CustomerName = 'Sam'
    AND Date = '04/29/15'

如果你想获得每个日期的count，你可以使用这个

SELECT T.Date,((ISNULL(CustomerCount,0) * 100) / TotalCount)
FROM 
(
SELECT COUNT(*) as TotalCount,Date 
FROM Customer 
WHERE CustomerName <> ''
GROUP BY Date
)T
LEFT JOIN 
(
SELECT COUNT(*)
FROM Customer
WHERE CustomerName = 'Sam'
GROUP BY Date 
)C
ON T.Date = C.Date

Answer 2

你可以试试这个：

select name, date, (100.0 * count(*)) / 
    (select count(*) from table t2 
     where name is not null and name <> '' and t2.date = t1.date) 
from table t1
where name is not null and name <> ''
group by name, date

Answer 3

对于所有用户

SELECT 
        (COUNT(*) * 100) / (SELECT COUNT(*) FROM Customer WHERE CustomerName <> '') as [percent],
         CustomerName
    FROM Customer group by CustomerName

针对特定用户

SELECT 
    (COUNT(*) * 100) / (SELECT COUNT(*) FROM Customer WHERE CustomerName <> '') as [percent],
     CustomerName
FROM Customer where CustomerName ='Sam'

Answer 4

您可以在子查询中计算每人+天的人数：

select  Date
,       CustomerName
,       100.0 * cnt / sum(cnt) over (partition by date)
from    (
        select  Date
        ,       CustomerName
        ,       count(*) cnt
        from    table1
        where   CustomerName <> ''
        group by
                Date
        ,       CustomerName
        ) t1

这会打印：

Date                    CustomerName 
----------------------- ------------ ---------------------------------------
2015-04-29 00:00:00.000 Joy          20.000000000000
2015-04-29 00:00:00.000 Oly          20.000000000000
2015-04-29 00:00:00.000 Sam          50.000000000000
2015-04-29 00:00:00.000 Tom          10.000000000000

(4 row(s) affected)

Answer 5

似乎每个人都在使用子查询或派生表。这应该表现良好并且很容易遵循。试试看：

DECLARE @CustomerName VARCHAR(5) = 'Sam';

SELECT  [Date],
        @CustomerName AS CustomerName,
        percentage = CAST(CAST(100.0 * SUM(CASE WHEN CustomerName = @CustomerName THEN 1 ELSE 0 END)/COUNT(*) AS INT) AS VARCHAR(20)) + '%'
FROM @yourTable
WHERE CustomerName != ''
GROUP BY [Date]

结果：

Date       CustomerName percentage
---------- ------------ ---------------------
2015-04-29 Sam          50%
2015-04-30 Sam          0%
2015-05-01 Sam          0%

如何根据列值获得百分比？

how to get a percentage depending on a column value?

sql

tsql

sql-server-2008-r2