代码之家 › 专栏 › 技术社区 › Jrgns

如何使用SQL计算树中的值之和

tree sql

Jrgns · 技术社区 · 16 年前

我需要对用户树所获得的每个级别的点数进行相加。级别1是低于用户级别1的用户的积分之和。二级是用户的一级点,低于用户二级等。。。

SQL看起来像什么?

如果你困惑了,别担心,我也是!

用户表:

ID    ParentID    Points
1     0           230
2     1           150
3     0           80
4     1           110
5     4           54
6     4           342

Tree:
0
|---\
1    3
| \
2  4---
    \  \
     5  6

输出应为:

ID    Points    Level1     Level2
1     230       150+110    150+110+54+342
2     150
3     80
4     110       54+342
5     54
6     342

最好是SQL Server语法和函数。。。

9 回复 | 直到 16 年前

Manrico Corazzi 16 年前

如果您使用的是Oracle DBMS,这将非常简单,因为Oracle支持使用语法。对于SQL Server,我想您可能会发现 Common Table Expressions 有用的

C B dkretz 7 年前

树不能很好地与SQL一起工作。如果您有非常(非常)少的写访问,您可以将树实现更改为使用嵌套集,这将使此查询非常容易。

SELECT SUM(points) 
FROM users 
where left > x and right < y

但是,树上的任何更改都需要接触大量行。最好在客户端中进行递归。

Grad van Horck 16 年前

或者如果您有最大数量的级别,您可以创建子查询,但它们的性能非常差。

(或者您可以获取mssqlserver2008并获取新的层次结构函数。。。;) )

tzot 16 年前

像其他人所说的,SQL通常不能很好地处理这种关系。通常,需要一个代理项“relations”表(id,parent_id,unique key on(id,parent_id)),其中:

INSERT INTO relations (id, parent_id) VALUES ([current_id], [current_id]);

INSERT INTO relations (id, parent_id) VALUES ([current_id], [current_parent_id]);

INSERT INTO relations (id, parent_id) SELECT [current_id], parent_id FROM relations WHERE id = [current_parent_id];
有逻辑避免循环
确保用存储过程处理“relations”上的更新和删除

有了那张桌子,你需要:

SELECT rel.parent_id, SUM(tbl.points)
FROM table tbl INNER JOIN relations rel ON tbl.id=rel.id
WHERE rel.parent_id <> 0
GROUP BY rel.parent_id;

Darrel Miller 16 年前

好吧,这给了你你想要的结果,但不能保证我没有错过什么。把它当作一个起点。我用了SQL 2005,SQL 2000不支持CTE

WITH Parent (id, GrandParentId, parentId, Points, Level1Points, Level2Points)
AS
(
    -- Find root
    SELECT id,  
            0 AS GrandParentId,
            ParentId,
            Points,
            0 AS Level1Points,
            0 AS Level2Points
    FROM tblPoints ptr
    WHERE ptr.ParentId = 0

    UNION ALL (
    -- Level2 Points
    SELECT pa.GrandParentId AS Id,
            NULL AS GrandParentId,
            NULL AS ParentId,
            0 AS Points, 
            0 AS Level1Points,
            pa.Points  AS Level2Points
    FROM tblPoints pt
            JOIN Parent pa ON pa.GrandParentId = pt.Id 
    UNION  ALL
    -- Level1 Points
    SELECT pt.ParentId AS Id,
            NULL AS GrandParentId,
            NULL AS ParentId,
            0 AS Points, 
            pt.Points AS Level1Points,
            0 AS Level2Points
    FROM tblPoints pt
            JOIN Parent pa ON pa.Id = pt.ParentId AND pa.ParentId IS NOT NULL 
    UNION  ALL
    -- Points
    SELECT pt.id,
            pa.ParentId AS GrandParentId,
            pt.ParentId,
            pt.Points, 
            0 AS Level1Points,
            0 AS Level2Points
    FROM tblPoints pt
            JOIN Parent pa ON pa.Id = pt.ParentId AND pa.ParentId IS NOT NULL )
)
SELECT id, 
    SUM(Points) AS Points,  
    SUM(Level1Points) AS Level1Points,
    CASE WHEN SUM(Level2Points) > 0 THEN  SUM(Level1Points) + SUM(Level2Points) ELSE 0 END AS Level2Points
FROM Parent
GROUP BY id 
ORDER by id

C B dkretz 7 年前

如果您正在处理存储在关系数据库中的树,我建议您查看“嵌套集”或“修改的预排序树遍历”。SQL将非常简单:

SELECT id, 
       SUM(value) AS value 
FROM table 
WHERE left>left\_value\_of\_your\_node 
  AND right<$right\_value\_of\_your\_node;

... 对你感兴趣的每个节点都这样做。

也许这能帮你: http://www.dbazine.com/oracle/or-articles/tropashko4 或者使用谷歌。

Ilya Kochetov 16 年前

使用游标和递归的用户定义函数调用(非常慢)
执行客户端递归计算(如果没有太多记录,则更可取)

Milan BabuÅ¡kov 16 年前

CREATE FUNCTION CALC
(
@node integer,
)
returns 
(
@total integer
)
as
begin
    select @total = (select node_value from yourtable where node_id = @node);

    declare @children table (value integer);
    insert into @children   
    select calc(node_id) from yourtable where parent_id = @node;

    @current = @current + select sum(value) from @children;
    return
end

Stef Heyenrath Dariusz WoÅºniak 14 年前

下表:

Id   ParentId
1   NULL
11    1
12    1
110 11
111 11
112 11
120 12
121 12
122 12
123 12
124 12

以及以下金额表:

只有叶(最后一级)Id定义了值。获取数据的SQL查询如下所示:

;WITH Data (Id, Val) AS
(
    select t.Id, SUM(v.val) as Val from dbo.TestTable t
    join dbo.Amount v on t.Id = v.Id
    group by t.Id
)

select cd.Id, ISNULL(SUM(cd.Val), 0) as Amount FROM
(
    -- level 3
    select t.Id, d.val from TestTable t
    left join Data d on d.id = t.Id

    UNION

    -- level 2
    select t.parentId as Id, sum(y.Val) from TestTable t
    left join Data y on y.id = t.Id
    where t.parentId is not null
    group by t.parentId

    UNION

    -- level 1
    select t.parentId as Id, sum(y.Val) from TestTable t
    join TestTable c on c.parentId = t.Id
    left join Data y on y.id = c.Id
    where t.parentId is not null
    group by t.parentId
) AS cd
group by id

Id     Amount
1     333555
11   555
12   333000
110 500
111 50
112 5
120 3000
121 30000
122 300000
123 0
124 0

我希望这能有帮助。