Den*_*s G 21 sql sql-server hierarchy hierarchical-data
对于简单的数据结构,例如:
ID parentID Text Price
1 Root
2 1 Flowers
3 1 Electro
4 2 Rose 10
5 2 Violet 5
6 4 Red Rose 12
7 3 Television 100
8 3 Radio 70
9 8 Webradio 90
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作为参考,层次结构树如下所示:
ID Text Price
1 Root
|2 Flowers
|-4 Rose 10
| |-6 Red Rose 12
|-5 Violet 5
|3 Electro
|-7 Television 100
|-8 Radio 70
|-9 Webradio 90
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我想计算每个级别的孩子数量.所以我会得到一个新的专栏"NoOfChildren",如下所示:
ID parentID Text Price NoOfChildren
1 Root 8
2 1 Flowers 3
3 1 Electro 3
4 2 Rose 10 1
5 2 Violet 5 0
6 4 Red Rose 12 0
7 3 Television 100 0
8 3 Radio 70 1
9 8 Webradio 90 0
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我读了一些关于分层数据的事情,但我不知何故被困在parentID上的多个内连接上.也许有人可以帮助我.
Lie*_*ers 25
使用CTE可以得到你想要的东西.
COUNT 每个根的项目.JOIN 这些再次与您的原始表格产生结果.测试数据
DECLARE @Data TABLE (
ID INTEGER PRIMARY KEY
, ParentID INTEGER
, Text VARCHAR(32)
, Price INTEGER
)
INSERT INTO @Data
SELECT 1, Null, 'Root', NULL
UNION ALL SELECT 2, 1, 'Flowers', NULL
UNION ALL SELECT 3, 1, 'Electro', NULL
UNION ALL SELECT 4, 2, 'Rose', 10
UNION ALL SELECT 5, 2, 'Violet', 5
UNION ALL SELECT 6, 4, 'Red Rose', 12
UNION ALL SELECT 7, 3, 'Television', 100
UNION ALL SELECT 8, 3, 'Radio', 70
UNION ALL SELECT 9, 8, 'Webradio', 90
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SQL语句
;WITH ChildrenCTE AS (
SELECT RootID = ID, ID
FROM @Data
UNION ALL
SELECT cte.RootID, d.ID
FROM ChildrenCTE cte
INNER JOIN @Data d ON d.ParentID = cte.ID
)
SELECT d.ID, d.ParentID, d.Text, d.Price, cnt.Children
FROM @Data d
INNER JOIN (
SELECT ID = RootID, Children = COUNT(*) - 1
FROM ChildrenCTE
GROUP BY RootID
) cnt ON cnt.ID = d.ID
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小智 5
考虑使用修改的预订树遍历方式来存储分层数据.请访问http://www.sitepoint.com/hierarchical-data-database/
确定任何节点的子节点数然后变得简单:
SELECT (right-left-1) / 2 AS num_children FROM ...
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