0

我有下表：

parent_id   child_id    child_class
1   2   1
1   3   1
1   4   2
2   5   2
2   6   2

Parent_id 表示文件夹 ID。子 ID 表示子文件夹（child_class=1）或子文件（child_class=2）。

我想通过以下方式仅获取所有文件（child_class = 2）的汇总计数器（自下而上）。例如，如果 C 是包含 5 个文件的叶文件夹（无子文件夹），而 B 是 C 的父文件夹，其中包含 4 个文件，则 C 上的计数器应为 5，B 上的计数器应为 9（=5从 C 加上 B 中的 4 个文件）等等递归地自下而上考虑兄弟文件夹等。

在上面的示例中，我希望得到以下结果（注意 3 是一个子文件夹，其中没有文件）：

parent_id   FilesCounter
3   0
2   2
1   3

我更喜欢 SQL 查询以提高性能，但功能也是可能的。

我尝试将分层查询与汇总（sql 2008 r2）混合，但到目前为止没有成功。

请指教。

4

2 回答 2

0

这个 CTE 应该可以解决问题......这是SQLFiddle。

SELECT parent_id, child_id, child_class,
(SELECT COUNT(*) FROM tbl a WHERE a.parent_id = e.parent_id AND child_class <> 1) AS child_count
INTO tbl2
FROM tbl e

;WITH CTE (parent_id, child_id, child_class, child_count)
AS
(
-- Start with leaf nodes
   SELECT parent_id, child_id, child_class, child_count 
   FROM tbl2
   WHERE child_id NOT IN (SELECT parent_id from tbl)
   UNION ALL
-- Recursively go up the chain
   SELECT e.parent_id, e.child_id, e.child_class, e.child_count + d.child_count
   FROM tbl2 e
   INNER JOIN CTE AS d
   ON e.child_id = d.parent_id
)
-- Statement that executes the CTE
SELECT FOLDERS.parent_id, max(ISNULL(child_count,0)) FilesCounter
FROM (SELECT parent_id FROM tbl2 WHERE parent_id NOT IN (select child_id from tbl2)
     UNION
     SELECT child_id FROM tbl2 WHERE child_class = 1) FOLDERS
LEFT JOIN CTE ON FOLDERS.parent_id = CTE.parent_id
GROUP BY FOLDERS.parent_id
于 2014-04-28T00:25:30.357 回答
0

Zak 的回答很接近，但根文件夹没有很好地汇总。以下工作：

with par_child as (
select 1 as parent_id,             2 as child_id,              1 as child_class
union all select 1,              3,              1
union all select 1,              4,              2
union all select 2,              5,              1
union all select 2,              6,              2
union all select 2,              10,           2  
union all select 3,              11,           2  
union all select 3,              7 ,             2
union all select 5,              8 ,             2
union all select 5,              9 ,             2
union all select 5,              12,           1  
union all select 5,              13,           1  
)
, child_cnt as 
(
      select parent_id as root_parent_id, parent_id, child_id, child_class, 1 as lvl from par_child    union all
      select cc.root_parent_id, pc.parent_id, pc.child_id, pc.child_class, cc.lvl + 1 as lvl from
      par_child pc join child_cnt cc on (pc.parent_id=cc.child_id)
),
distinct_folders as (
select distinct child_id as folder_id from par_child where child_class=1
)
select root_parent_id, count(child_id) as cnt from child_cnt where child_class=2 group by root_parent_id
union all
select folder_id, 0 from distinct_folders df where not exists (select 1 from par_child pc where df.folder_id=pc.parent_id)
于 2014-04-29T01:38:26.477 回答