Question

我有一个表，其中包含1个名为Week的列和5个名为n1，n2，n3，n4和n5的列。我想计算每周有多少n1，n2 ... n5列匹配。

到目前为止，我创建了一个查询，显示按周排序的所有n1 ... n5

select e.Week, e.Year, e.n1, e.n2, e.n3, e.n4, e.n5, e.Date
from Entries e
order by e.Week asc, e.Year asc

表

CREATE TABLE [dbo].[Enties](
    [Id] [int] IDENTITY(1,1) NOT NULL,  
    [Week] [int] NULL,
    [n1] [int] NULL,
    [n2] [int] NULL,
    [n3] [int] NULL,
    [n4] [int] NULL,
    [n5] [int] NULL,
    [sup] [int] NULL
 CONSTRAINT [PK_Enties] PRIMARY KEY CLUSTERED 
(
    [Id] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]

我使用的是SQL Server，但这是一个常见的SQL问题。任何想法我怎样才能做到这一点？

更新：假设该表包含3年，每年3个相等的星期。我想知道有多少

week1.n1=week2.n1=week3.n1, week1.n2=week2.n2=week3.n2, ...

每周都相等。

结果将类似于

Week1 = 0 match,
Week2 = 2 match,
Week3 = 1 matches,
...

Answer 1

在SQL Server中，如果在CASE表达式中包装，则可以使用布尔表达式作为其数值。

您可以计算同时匹配的数量：

select
    (Round/2) as Week
,   e.Year
,   SUM(
        (case when n1=n2 then 1 else 0 end)
    +   (case when n1=n3 then 1 else 0 end)
    +   (case when n1=n4 then 1 else 0 end)
    +   (case when n1=n5 then 1 else 0 end)
    +   (case when n2=n3 then 1 else 0 end)
    +   (case when n2=n4 then 1 else 0 end)
    +   (case when n2=n5 then 1 else 0 end)
    +   (case when n3=n4 then 1 else 0 end)
    +   (case when n3=n5 then 1 else 0 end)
    +   (case when n4=n5 then 1 else 0 end)
    ) as MatchCount
from Entries
group by (Round/2), Year
order by Round asc, Year asc

SUM的参数列出了表中所有{n _i，n _j}的对，计算成对匹配的数量。

计算几年中同一周的相等列数

1 个答案: