Question

好的，这个问题有点难以解释，但在这里......

我有一个查询，它可以提取在树中呈现的数据，就像可视化一样。在此查询中计算的yloc用于确定y位置以呈现附加到这些功能的某些对象。

仅使用MAX（合并（CAP_TREE2.depth），0）不起作用。所以我写了一个算法，在某些情况下显示我需要的是什么。

以下是算法：

    if (capability has c2 !=1)
    {
        yloc = 0
    }
    else if (capability is child (at any depth) of capability with c2 != 1)
    {
        yloc = depth - (depth of first parent with c2 != 1)
    }
    else
    {
        y = depth
    }

注意： c2代表childNumber（按字母顺序排列）。因此，如果一个能力有两个孩子“A”和“B”。 “A”将具有c2 = 1并且“B”将具有c2 = 2。

我在尝试在SQL（Oracle 11g R2）中实现此算法时遇到了很多麻烦。我在完成问题时遇到的问题是我的算法中的 else if 子句。我已经能够通过c2！= 1获得所有功能的孩子（在任何深度）。

然而，我无法完成的部分是 yloc =深度 - （第一个父亲的深度为c2！= 1）

任何人都可以告诉我如何获得“c2的第一个父亲的深度！= 1”？如果我能得到那个值，我的查询最终会完成。

我将展示我希望从我的样本数据中获得的值，以便更加清晰。

“投资管理”第一个父亲与c2！= 1是“FPP”，其深度为2。具有c2！= 1的“第三方产品管理”第一父母是深度为2的“FPP”。 “订单管理（PI）”第一个父项c2！= 1是“操作和服务（PI）”，深度为2。

注意：抱歉，我无法更改架构。

这是我到目前为止所做的...最底层的选择陈述是我工作的地方。 http://sqlfiddle.com/#!4/55b5a/116

架构SQL：

CREATE TABLE capability 
    (
     id int,
     parent_id int,
     name varchar(200)
    )
;

CREATE TABLE tree
  (
     descendantid int,
     ancestorid int,
     depth int
  )
;

INSERT INTO capability (id, parent_id, name) VALUES (1, -1, 'BU-Specific Capabilities');
INSERT INTO capability (id, parent_id, name) VALUES (2, 1, 'PI Capability Model');
INSERT INTO capability (id, parent_id, name) VALUES (3, 2, 'Core Business Processing (PI)');
INSERT INTO capability (id, parent_id, name) VALUES (4, 3, 'Institutional Trust Administration (PI)');
INSERT INTO capability (id, parent_id, name) VALUES (5, 2, 'FPP');
INSERT INTO capability (id, parent_id, name) VALUES (6, 5, 'Investment Management');
INSERT INTO capability (id, parent_id, name) VALUES (7, 6, '3rd party Product Management');
INSERT INTO capability (id, parent_id, name) VALUES (8, 2, 'Operations and Shared Services (PI)');
INSERT INTO capability (id, parent_id, name) VALUES (9, 8, 'Order Management (PI)');
INSERT INTO capability (id, parent_id, name) VALUES (10, 8, 'Settlements (PI)');
INSERT INTO capability (id, parent_id, name) VALUES (11, -1, 'Common Core Capabilities');

INSERT INTO tree (descendantid, ancestorid, depth) VALUES (1, 1, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (2, 1, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (2, 2, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (3, 1, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (3, 2, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (3, 3, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (4, 1, 3);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (4, 2, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (4, 3, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (4, 4, 0);                                               
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (5, 1, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (5, 2, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (5, 5, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (6, 1, 3);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (6, 2, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (6, 5, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (6, 6, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (7, 1, 4);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (7, 2, 3);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (7, 5, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (7, 6, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (7, 7, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (8, 1, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (8, 2, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (8, 8, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (9, 1, 3);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (9, 2, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (9, 8, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (9, 9, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (10, 1, 3);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (10, 2, 2);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (10, 8, 1);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (10, 10, 0);
INSERT INTO tree (descendantid, ancestorid, depth) VALUES (11, 11, 0);

查询SQL：

SELECT
  cap.name,
  max(coalesce(CAP_TREE2.depth,0)) as ydepth,

  CASE
    WHEN cap.parent_id != -1
       THEN DENSE_RANK() OVER (PARTITION BY cap.parent_id ORDER BY cap.name) --child number
    ELSE
       1
  END as c2

FROM capability cap
    INNER JOIN tree CAP_TREE2 ON CAP.id = CAP_TREE2.descendantid

group by cap.name, cap.id, cap.parent_id
;


  WITH ids(id, c2) AS ( -- Find the ids and the ranks
  SELECT
    id, 
    CASE
    WHEN parent_id = -1
      THEN 1
    ELSE
      DENSE_RANK() OVER (PARTITION BY parent_id ORDER BY name) --child number
    END as c2
  FROM capability
),
t(id, depth) AS (
    SELECT id, 0 AS depth FROM ids WHERE c2 != 1  -- Take only ranks not equal to one
UNION ALL
    SELECT c.id, depth+1 FROM capability c JOIN t ON (c.parent_id = t.id)  -- Tree-walking
)
--SELECT DISTINCT cap.name, t.depth
--SELECT DISTINCT cap.name, min(t.depth)
--FROM capability cap JOIN t USING(id)
--group by cap.name


SELECT
  cap.name,
  CASE

    -- cap has childNumber != 1
    WHEN DENSE_RANK() OVER (PARTITION BY cap.parent_id ORDER BY cap.name) != 1
      THEN 0 -- y = 0

    -- cap is child of cap with childNumber != 1
    WHEN cap.name IN (SELECT DISTINCT capa.name
                      FROM capability capa
                      JOIN t USING(id))
         THEN max(coalesce(CAP_TREE2.depth,0)) -- y = depth - (depth of first parent with childNumber != 1)

    ELSE max(coalesce(CAP_TREE2.depth,0)) -- y = depth 

  END as yloc


FROM capability cap
    INNER JOIN tree CAP_TREE2 ON CAP.id = CAP_TREE2.descendantid

group by cap.name,
         cap.id,
         cap.parent_id

此外，这是我想要的结果集

NAME                                            DEPTH               C2            YLOC

BU-Specific Capabilities                          0                 1               0       --yloc=depth
Common Core Capabilities                          0                 1               0       --yloc=depth
PI Capability Model                               1                 1               1       --yloc=depth
Core Business Processing (PI)                     2                 1               2       --yloc=depth
FPP                                               2                 2               0       --yloc=0
Operations and Shared Services (PI)               2                 3               0       --yloc=0
Institutional Trust Administration (PI)           3                 1               3       --yloc=depth
Investment Management                             3                 1               1       --yloc=depth - (depth of FPP)
3rd party Product Management                      4                 1               2       --yloc=depth - (depth of FPP)
Order Management (PI)                             3                 1               1       --yloc=depth -(depth of Operations and..)
Settlements (PI)                                  3                 2               0       --yloc=0

Answer 1

我认为您不需要树表，因为您已经在功能表中声明了从父级到子级的关系。

当您使用oracle中的 CONNECT BY PRIOR 命令时，它会为您构建树，并允许您使用 LEVEL 伪列来确定节点的深度。 / p>

唯一棘手的部分是获得深度为＆gt;的第一个父节点。据我所知，你的问题应该是c2。结果在列中的 FIRST_PARENT_IN_DEPTH_G_1 即可。

SELECT ID
    , PARENT_ID
    , NAME
    , DEPTH
    , NODES_PER_LEVEL
    , TO_NUMBER(SUBSTR(ID_PATH, LENGTH(REGEXP_SUBSTR(ID_PATH, '^\s\d{1,}\s')) + 1, 
            INSTR(SUBSTR(ID_PATH, LENGTH(REGEXP_SUBSTR(ID_PATH, '^\s\d{1,}\s')) + 1), ' '))) 
            AS FIRST_PARENT_IN_DEPTH_G_1
FROM (
    SELECT ID
        , PARENT_ID
        , NAME
        , LEVEL DEPTH
        , COUNT(*) OVER (PARTITION BY PARENT_ID, LEVEL) NODES_PER_LEVEL
        , sys_connect_by_path( id, ' ' ) ID_PATH
    FROM capability 
    START WITH PARENT_ID = -1
    CONNECT BY PRIOR ID = PARENT_ID
    )    
ORDER BY ID_PATH

SQL物化路径还是有更简单的方法吗？

1 个答案: