分组，多个计数，多个表SQL Server

Question

我有两个表，如下所示：

第一张桌子：2016年

Os          |   Count
------------+----------
Windows 7         8
Windows 7         9
Windows 7        20
Windows 8        30
Linux            15

第二表：2017年

Os          |   Count
------------+----------
Windows 7        35
Windows 7        11
Windows 8        10
Windows 8         8
Linux            10
Ubuntu            3

我自己尝试，但是我的挑战是在同时计算两个不同表中的两个字段时使用GROUP BY函数，但总是出错。

先感谢您，

更新：

抱歉，我在解释我的请求时犯了一个错误：我需要写一个查询来获得如下结果：

    --OS--         --2016--    --2017--
    ------------------------------------
    Windows 7        37             46
    Windows 8        30             18
    Linux            15             10
    Ubuntu           0              3

Answer 1

您可以按以下方式进行简单分组：

select a.OS, sum(count) from (
    select * from your2016table
        union all 
    select * from your2017table
    ) a
    group by a.OS

为此，您可以将year作为分组依据并进行透视

;with cte as (
select a.OS, a.[year], SumCt = sum(count) from (
    select *, 2016 as [year] from your2016table
        union all 
    select *, 2017 as [year] from your2017table
    ) a
    group by a.OS, a.[year]
)
select * from cte 
pivot (max(sumct) for [year] in ([2016], [2017])) p

Answer 2

没有UNION ALL您可以执行以下操作：

SELECT
    COALESCE( [2016].[OS], [2017].[OS] ) AS [OS],
    ( [2016].[Count] + [2017].[Count] ) AS [Count]
FROM
    [2016]
    FULL OUTER JOIN [2017] ON [2016].[OS] = [2017].[OS]

假设OS列不包含重复项，这也消除了对GROUP BY的需要。

需要通过GROUP BY （而不是SELECT DISTINCT ）消除重复项，但是适用相同的结构：

SELECT
    COALESCE( [2016-D].[OS], [2017].[OS] ) AS [OS],
    ( [2016-D].[Count] + [2017].[Count] ) AS [Count]
FROM
    (
        SELECT [OS], SUM( [Count] ) FROM [2016] GROUP BY [OS]
    ) AS [2016-D]
    FULL OUTER JOIN
    (
        SELECT [OS], SUM( [Count] ) FROM [2017] GROUP BY [OS]
    ) AS [2017-D]
        ON [2016].[OS] = [2017].[OS]

...但是随后变得非常笨拙，并且UNION ALL方法变得更加简单！

更新：

OP修改了他们的问题，说他们想要单独的列-因此在这里使用OUTER JOIN成为理想的解决方案：

SELECT
    COALESCE( [2016-D].[OS], [2017-D].[OS] ) AS [OS],
    [2016-D].[Count] AS [2016],
    [2017-D].[Count] AS [2017]
FROM
    (
        SELECT
            [OS],
            SUM( [Count] ) AS [Count]
        FROM
            [2016]
        GROUP BY
            [OS]
    ) AS [2016-D]
    FULL OUTER JOIN
    (
        SELECT
            [OS],
            SUM( [Count] ) AS [Count]
        FROM
            [2017]
        GROUP BY
            [OS]
    ) AS [2017-D]
        ON [2016].[OS] = [2017].[OS]
ORDER BY
    [OS]