Question

我有两张表如下：

第一张表：2016

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

第二张表：2017

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

我自己尝试，但我的挑战是使用GROUP BY功能，同时我计算两个不同表中的两个字段，我总是错误。

提前谢谢你，

更新

对不起，我在解释我的请求时犯了一个错误：我需要写一个查询来得到这样的结果：

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

Answer 1

你可以通过以下方式做一个简单的小组：

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

为此你可以使用year作为group by和pivot

;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

如果没有$ python -c "import vtk; print vtk.vtkSphereSource()" vtkSphereSource (0xcfdcc0) Debug: Off Modified Time: 39 Reference Count: 2 Registered Events: (none) Executive: 0xae5e00 ErrorCode: Success Information: 0xf6e210 AbortExecute: Off Progress: 0 Progress Text: (None) Theta Resolution: 8 Phi Resolution: 8 Theta Start: 0 Phi Start: 0 Theta End: 360 Phi End: 180 Radius: 0.5 Center: (0, 0, 0) LatLong Tessellation: 0 Output Points Precision: 0，您可以执行此操作：

UNION ALL

这也消除了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]的需要，假设GROUP BY列不包含重复项。

需要通过OS（而不是GROUP BY）删除重复项，但适用相同的结构：

SELECT DISTINCT

...但是它变得非常笨拙，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]方法变得更简单了！

更新

OP修改了他们的问题，说他们想要单独的列 - 所以这就是使用UNION ALL成为理想解决方案的地方：

OUTER JOIN

分组依据，多个计数，多个表SQL Server

2 个答案:

更新