2

我在 SQL Server 2008 R2 中有两个表：

Store:                      Record:
--------------              ------------------------
Shape       ID              PickedShape     Finished
Circle      1               Circle          Y
Circle      2               Circle          N
Circle      3               Square          N
Square      1               Square          N
Square      2               Oval            Y
Triangle    1               Oval            Y

我想编写一个 SQL 查询来输出两个表中所有可能的形状及其计数。我可以用愚蠢的方式写它，比如

select 
    'Circle', (select count(1) from Store where Shape = 'Circle'), 
    (select count(1) from Record where PickedShape = 'Circle'), 
    (select count(1) from Record where Finished = 'Y' and PickedShape = 'Circle') 
UNION
select 
    'Square', (select count(1) from Store where Shape = 'Square'), 
    (select count(1) from Record where PickedShape = 'Square'), 
    (select count(1) from Record where Finished = 'Y' and PickedShape = 'Square') 
UNION...

继续，但这只是愚蠢而且效率不高。

我认为更聪明的方法是使用 group by。由于有些人可能不喜欢用勺子喂别人，所以这是我尝试过的

SELECT 
    Shape, COUNT(Shape) AS Available, Picked, Finished
FROM 
    Store 
FULL JOIN 
    (SELECT PickedShape, COUNT(1) As Picked, SUM(CASE WHEN Finished='Y' THEN 1 ELSE 0 END) AS Finished 
     FROM Record 
     GROUP BY PickedShape) t2 ON Store.Shape = t2.PickedShape
GROUP BY 
    Shape, Picked, Finished

输出是

Shape   Available   Picked  Finished
NULL        0       2       2
Circle      3       2       1
Square      2       2       0
Triangle    1       NULL    NULL

你可以看到问题出在哪里。

首先，我想在 Shape 下使用“Oval”而不是 NULL。使用 FULL JOIN 为我提供了两个表中的所有变体，但没有显示它们......

其次，我希望 Picked and Finished 为缺失的显示 0 而不是 NULL。

第三，如果可能的话，我希望 SQL 更高效。

你如何解决这些问题？

谢谢！

4

1 回答 1

1

我稍微修改了您的查询在您按形状分组的第一个查询中，但第一个查询中不存在“椭圆”。因此 Shape 从派生查询中接收。派生查询中不存在三角形，因此其数量等于 0。

SELECT 
  COALESCE(Shape, t2.PickedShape) AS Shape, COUNT(Shape) AS Available,
  ISNULL(Picked, 0) AS Picked, ISNULL(Finished, 0) AS Finished
FROM 
 Store 
FULL JOIN 
  (SELECT PickedShape, COUNT(1) As Picked, SUM(CASE WHEN Finished='Y' THEN 1 ELSE 0 END) AS Finished 
   FROM Record 
   GROUP BY PickedShape) t2 ON Store.Shape = t2.PickedShape
GROUP BY 
  Shape, t2.PickedShape, Picked, Finished
于 2012-11-09T07:41:50.163 回答