1

假设我有一张桌子TableA

parent_id int
this_id int
filter int
this_date date

以及一些示例数据:

parent_id = 1, this_id = 1, filter = 1, this_date = ...
parent_id = 1, this_id = 2, filter = 0, this_date = ...
parent_id = 1, this_id = 3, filter = 1, this_date = ...
parent_id = 4, this_id = 4, filter = 0, this_date = ...
parent_id = 4, this_id = 5, filter = 0, this_date = ...
parent_id = 4, this_id = 6, filter = 1, this_date = ...
parent_id = null, this_id = 7, filter = 0, this_date = ...
parent_id = null, this_id = 8, filter = 1, this_date = ...

parent_id总是和其中一个孩子的一样this_id。如果没有孩子,parent_id则为空

桌子不能改变,这就是我们所拥有的。

我创建了一个视图来给我总结:

SELECT ISNULL(parent_id,this_id) id, COUNT(*) numparts, MAX(this_date)...
FROM TableA
GROUP BY ISNULL(parent_id,this_id)

我想在我的摘要视图中添加具有filter=1.

所以我对这个例子的结果(如果我做 a SELECT * FROM theview)将是:

id  numparts dt    numOfFilter1
1      3     ...        2
4      3     ...        1
7      1     ...        0
8      1     ...        1

实际上,我的过滤器列将是一个子查询,但我想一旦我对这个位进行了排序,我就可以弄清楚这一点。

4

1 回答 1

1

试试这样:

SELECT  ISNULL(parent_id,this_id) id, 
        COUNT(*) numparts, 
        MAX(this_date), 
        SUM(CASE WHEN filter = 1 THEN 1 ELSE 0 END) numOfFilter1
FROM    TableA
GROUP BY 
        ISNULL(parent_id,this_id)
于 2012-12-07T10:35:28.267 回答