树结构中的递归和

Question

我在一张桌子上有一个树木结构.该表是一个可以无限嵌套的类别树.每个类别都有一个ProductCount列,用于说明该类别中有多少产品(不是对子类别求和).

Id  | ParentId | Name      | ProductCount
------------------------------------
1   | -1       | Cars      | 0
2   | -1       | Bikes     | 1
3   | 1        | Ford      | 10
4   | 3        | Mustang   | 7
5   | 3        | Focus     | 4

我想做一个SQL查询,每个行/类别给我产品的数量,包括子类别中的产品.

上表的输出应该是

Id  | ParentId | Name      | ProductCount | ProductCountIncludingChildren
--------------------------------------------------------------------------
1   | -1       | Cars      | 0            | 21
2   | -1       | Bikes     | 1            | 1
3   | 1        | Ford      | 10           | 21
4   | 3        | Mustang   | 7            | 7
5   | 3        | Focus     | 4            | 4

我知道我可能应该使用CTE,但不能让它以它应该的方式工作.

任何帮助表示赞赏!

Answer 1

您可以使用递归CTE,其中您在锚点部分获取所有行,并在递归部分连接以获取子行.记住锚点部分的原始Id别名RootID,并在分组的主查询中进行求和RootID.

SQL小提琴

MS SQL Server 2012架构设置:

create table T
(
  Id int primary key,
  ParentId int,
  Name varchar(10),
  ProductCount int
);

insert into T values
(1, -1, 'Cars',    0),
(2, -1, 'Bikes',   1),
(3,  1, 'Ford',    10),
(4,  3, 'Mustang', 7),
(5,  3, 'Focus',   4);

create index IX_T_ParentID on T(ParentID) include(ProductCount, Id);

查询1:

with C as
(
  select T.Id,
         T.ProductCount,
         T.Id as RootID
  from T
  union all
  select T.Id,
         T.ProductCount,
         C.RootID
  from T
    inner join C 
      on T.ParentId = C.Id
)
select T.Id,
       T.ParentId,
       T.Name,
       T.ProductCount,
       S.ProductCountIncludingChildren
from T
  inner join (
             select RootID,
                    sum(ProductCount) as ProductCountIncludingChildren
             from C
             group by RootID
             ) as S
    on T.Id = S.RootID
order by T.Id
option (maxrecursion 0)

结果:

| ID | PARENTID |    NAME | PRODUCTCOUNT | PRODUCTCOUNTINCLUDINGCHILDREN |
|----|----------|---------|--------------|-------------------------------|
|  1 |       -1 |    Cars |            0 |                            21 |
|  2 |       -1 |   Bikes |            1 |                             1 |
|  3 |        1 |    Ford |           10 |                            21 |
|  4 |        3 | Mustang |            7 |                             7 |
|  5 |        3 |   Focus |            4 |                             4 |

Answer 2

这与汤姆的答案是相同的概念，但代码更少（而且速度更快）。

with cte as
(
  select v.Id, v.ParentId, v.Name, v.ProductCount, 
  cast('/' + cast(v.Id as varchar) + '/' as varchar) Node
  from Vehicle v
  where ParentId = -1
  union all
  select v.Id, v.ParentId, v.Name, v.ProductCount,  
  cast(c.Node + CAST(v.Id as varchar) + '/' as varchar)
  from Vehicle v
  join cte c on v.ParentId = c.Id
)

select c1.Id, c1.ParentId, c1.Name, c1.ProductCount, 
c1.ProductCount + SUM(isnull(c2.ProductCount, 0)) ProductCountIncludingChildren
from cte c1
left outer join cte c2 on c1.Node <> c2.Node and left(c2.Node, LEN(c1.Node)) = c1.Node
group by c1.Id, c1.ParentId, c1.Name, c1.ProductCount
order by c1.Id

SQL Fiddle（我添加了一些额外的数据行进行测试）