一个多对多表

Question

递归关系，我成立了一个多对多递归表是这样的：

CREATE TABLE PROD 
(
IDPROD INT NOT NULL PRIMARY KEY, 
NAME VARCHAR(3) 
); 

CREATE TABLE COMP 
(
IDPARENT INT REFERENCES PROD(IDPROD), 
IDCHILD INT REFERENCES PROD(IDPROD) 
); 

INSERT INTO PROD (IDPROD, NAME) VALUES (1, 'abc'); 
INSERT INTO PROD (IDPROD, NAME) VALUES (2, 'def'); 
INSERT INTO PROD (IDPROD, NAME) VALUES (3, 'ghi'); 
INSERT INTO PROD (IDPROD, NAME) VALUES (4, 'jkl'); 
INSERT INTO PROD (IDPROD, NAME) VALUES (5, 'mno'); 

INSERT INTO COMP (IDPARENT, IDCHILD) VALUES (1, 2); 
INSERT INTO COMP (IDPARENT, IDCHILD) VALUES (3, 4); 
INSERT INTO COMP (IDPARENT, IDCHILD) VALUES (4, 5); 

随着recursivs CTE我可以从第二个表中的特定节点的所有孩子。

WITH RECURSIVE TEST (IDPARENT, IDCHILD) AS 
(SELECT P0.IDPARENT, P0.IDCHILD 
FROM COMP AS P0 
WHERE P0.IDPARENT = 3 
UNION ALL 
SELECT P1.IDPARENT, P1.IDCHILD 
FROM COMP AS P1, TEST AS T 
WHERE T.IDCHILD = P1.IDPARENT) 

SELECT * FROM TEST 

但我需要一个查询，这将使我的整个结构，而不仅仅是一个节点。就像在经典的邻接列表中，您可以获得IDPARENT为NULL的所有根节点以及下面列出的子节点。我使用火鸟。

Answer 1

我不熟悉的火鸟，但这部作品在SQL Server中，所以是希望类似/足以让您的轨道上：

WITH TEST (IDRoot, IDPARENT, IDCHILD) AS 
(

    SELECT P0.IDPROD, C0.IDParent, C0.IDCHILD 
    FROM PROD AS P0 
    left outer join COMP C0 on C0.IDParent = P0.IDPROD 
    WHERE P0.IDProd not in (select IDChild from COMP) 

    UNION ALL 

    SELECT T.IDRoot, C1.IDPARENT, C1.IDCHILD 
    FROM COMP AS C1 
    inner join TEST AS T on T.IDCHILD = C1.IDPARENT 

) 
SELECT * FROM TEST 

希望有所帮助。

SQL小提琴版本：http://sqlfiddle.com/#!6/22f84/7

注意

Includ列来表示树的根，以及父/子 - 因为可能有多个树，如果我们不指定特定的根：

WITH TEST (IDRoot, IDPARENT, IDCHILD) AS 

治疗任何产品，其是不是小孩作为根（即，第一项在树）。

WHERE P0.IDProd not in (select IDChild from COMP) 

编辑：回答任何节点以查看其所有的亲戚在评论

查询：

简单的方法来过滤任何节点上会修改上面的语句的WHERE P0.IDProd not in (select IDChild from COMP)与WHERE P0.IDProd = IdImInterestedIn。但是，如果您想使用CTE查看视图，则可以通过此静态查询运行查询，您可以使用下面的代码 - 然后，您可以在IDProd（select * from test where IDProd = IdImInterestedIn）上筛选以查看该项目的祖先和后代。

WITH TEST (IDProd, IDRelation, Generation) AS 
(

    SELECT IDPROD 
    , IDPROD 
    , 0 
    FROM PROD 

    UNION ALL 

    SELECT T.IDPROD 
    , C.IdParent 
    , T.Generation - 1 
    FROM TEST AS T 
    inner join Comp as C 
    on C.IdChild = T.IDRelation 
    where t.Generation <= 0 

    UNION ALL 

    SELECT T.IDPROD 
    , C.IdChild 
    , T.Generation + 1 
    FROM TEST AS T 
    inner join Comp as C 
    on C.IdParent = T.IDRelation 
    where t.Generation >= 0 

) 
SELECT * 
FROM TEST 
order by IDProd, Generation 

SQL小提琴：http://sqlfiddle.com/#!6/22f84/15

见根节点的全树一列

WITH TEST (IDRoot, IDPARENT, IDCHILD, TREE) AS 
(

    SELECT P0.IDPROD, C0.IDParent, C0.IDCHILD, cast(P0.IDPROD as nvarchar(max)) + coalesce(', ' + cast(C0.IDCHILD as nvarchar(max)),'') 
    FROM PROD AS P0 
    left outer join COMP C0 on C0.IDParent = P0.IDPROD 
    WHERE P0.IDProd not in (select IDChild from COMP) 

    UNION ALL 

    SELECT T.IDRoot, C1.IDPARENT, C1.IDCHILD, TREE + coalesce(', ' + cast(C1.IDCHILD as nvarchar(max)),'') 
    FROM COMP AS C1 
    inner join TEST AS T on T.IDCHILD = C1.IDPARENT 

) 
SELECT * 
FROM TEST 
order by IDRoot 

SQL小提琴：http://sqlfiddle.com/#!6/22f84/19

编辑：回答到其他意见

with cte (tree_root_no, tree_row_no, relation_sort, relation_chart, Name, id, avoid_circular_ref) as 
(
     select row_number() over (order by p.idprod) 
     , 1 
     , cast(row_number() over (order by p.idprod) as nvarchar(max)) 
     , cast('-' as nvarchar(max)) 
     , p.NAME 
     , p.IDPROD 
     , ',' + cast(p.IDPROD as nvarchar(max)) + ',' 
     from PROD p 
     where p.IDPROD not in (select IDCHILD from COMP) --if it's nothing's child, it's a tree's root 

     union all 

     select cte.tree_root_no 
     , cte.tree_row_no + 1 
     , cte.relation_sort + cast(row_number() over (order by p.idprod) as nvarchar(max)) 
     , replace(relation_chart,'-','|') + ' -' 
     , p.NAME 
     , p.IDPROD 
     , cte.avoid_circular_ref + cast(p.IDPROD as nvarchar(max)) + ',' 
     from cte 
     inner join COMP c on c.IDPARENT = cte.id 
     inner join PROD p on p.IDPROD = c.IDCHILD 
     where charindex(',' + cast(p.IDPROD as nvarchar(max)) + ',', cte.avoid_circular_ref) = 0 
) 
select tree_root_no, tree_row_no, relation_sort, relation_chart, id, name 
from cte 
order by tree_root_no, relation_sort 

SQL小提琴：http://sqlfiddle.com/#!6/4397f/9

更新，以显示每路

这一个是一个讨厌的黑客，但我能想到的唯一的办法来解决你的难题;这给每个路径穿过树林自己的号码：

;with inner_cte (parent, child, sequence, treePath) as (

    select null 
    , p.IDPROD 
    , 1 
    , ',' + CAST(p.idprod as nvarchar(max)) + ',' 
    from @prod p 
    where IDPROD not in 
    (
     select IDCHILD from @comp 
    ) 

    union all 

    select cte.child 
    , c.IDCHILD 
    , cte.sequence + 1 
    , cte.treePath + CAST(c.IDCHILD as nvarchar(max)) + ',' 
    from inner_cte cte 
    inner join @comp c on c.IDPARENT = cte.child 

) 
, outer_cte (id, value, pathNo, sequence, parent, treePath) as 
(
    select icte.child, p.NAME, ROW_NUMBER() over (order by icte.child), icte.sequence, icte.parent, icte.treePath 
    from inner_cte icte 
    inner join @prod p on p.IDPROD = icte.child 
    where icte.child not in (select coalesce(icte2.parent,-1) from inner_cte icte2) 

    union all 

    select icte.child, p.NAME, octe.pathNo,icte.sequence, icte.parent, icte.treePath 
    from outer_cte octe 
    inner join inner_cte icte on icte.child = octe.parent and CHARINDEX(icte.treePath, octe.treePath) > 0 
    inner join @prod p on p.IDPROD = icte.child 

) 
select id, value, pathNo 
from outer_cte 
order by pathNo, sequence 

SQL小提琴这里：http://sqlfiddle.com/#!6/5a16e/1