Question

表T表示树。每条记录都是一个节点，每个节点只有一个父节点。

此查询计算每个节点的每个分支的SUM（）。

WITH t AS
        (SELECT  1 id, NULL parent_id, NULL value FROM dual UNION ALL
         SELECT 10 id,    1 parent_id, 1000 value FROM dual UNION ALL
         SELECT 20 id,    1 parent_id, 2000 value FROM dual UNION ALL
         SELECT 30 id,   10 parent_id, 3000 value FROM dual UNION ALL
         SELECT 40 id,   10 parent_id, 4000 value FROM dual UNION ALL
         SELECT 50 id,   20 parent_id, 5000 value FROM dual UNION ALL
         SELECT 60 id,    1 parent_id, 6000 value FROM dual UNION ALL
         SELECT 70 id,   60 parent_id, 7000 value FROM dual UNION ALL
         SELECT 80 id,   70 parent_id, 8000 value FROM dual
    ) SELECT CAST(LPAD(' ', (LEVEL-1)*4) || ID AS VARCHAR2(20))  id
        ,VALUE                                                   self_value
        ,(SELECT SUM (value)
          FROM   t t2
          CONNECT BY 
            PRIOR t2.ID = t2.parent_id
            START WITH t2.id = t.id)                             branch_value
      FROM   t
      CONNECT BY PRIOR t.id = t.parent_id
      START WITH t.parent_id IS NULL
      ORDER SIBLINGS BY t.id;

ID                   SELF_VALUE BRANCH_VALUE
-------------------- ---------- ------------
1                                      36000
    10                     1000         8000
        30                 3000         3000
        40                 4000         4000
    20                     2000         7000
        50                 5000         5000
    60                     6000        21000
        70                 7000        15000
            80             8000         8000

9 rows selected.

我一直在尝试使用替代的Subquery Factoring语法来实现此查询的相同结果。任何帮助都会非常感激！

Answer 1

在递归查询中支持GROUP BY之前，我认为只有一个查询是不可能的。所以我在WITH - 子句中添加了第二个子查询。也许这足以解决你的问题（很好的脑筋急转弯）。

这将计算数据：

WITH t AS
        (SELECT  1 id, NULL parent_id, NULL value FROM dual UNION ALL
         SELECT 10 id,    1 parent_id, 1000 value FROM dual UNION ALL
         SELECT 20 id,    1 parent_id, 2000 value FROM dual UNION ALL
         SELECT 30 id,   10 parent_id, 3000 value FROM dual UNION ALL
         SELECT 40 id,   10 parent_id, 4000 value FROM dual UNION ALL
         SELECT 50 id,   20 parent_id, 5000 value FROM dual UNION ALL
         SELECT 60 id,    1 parent_id, 6000 value FROM dual UNION ALL
         SELECT 70 id,   60 parent_id, 7000 value FROM dual UNION ALL
         SELECT 80 id,   70 parent_id, 8000 value FROM dual),
hierarchy (id,ancestor,value) AS (
  SELECT t.id,t.id,t.value
    FROM t
  UNION ALL
  SELECT t.id,h.ancestor,t.value
    FROM t
      INNER JOIN hierarchy h
        ON t.parent_id = h.id)
SELECT h.ancestor, t.parent_id, t.value, SUM(h.value)
  FROM hierarchy h
    INNER JOIN t
      ON t.id = h.ancestor
  GROUP BY h.ancestor,t.value,t.parent_id;

要获得与问题中描述的相同的顺序和格式，请添加路径和深度的计算：

WITH t AS
        (SELECT  1 id, NULL parent_id, NULL value FROM dual UNION ALL
         SELECT 10 id,    1 parent_id, 1000 value FROM dual UNION ALL
         SELECT 20 id,    1 parent_id, 2000 value FROM dual UNION ALL
         SELECT 30 id,   10 parent_id, 3000 value FROM dual UNION ALL
         SELECT 40 id,   10 parent_id, 4000 value FROM dual UNION ALL
         SELECT 50 id,   20 parent_id, 5000 value FROM dual UNION ALL
         SELECT 60 id,    1 parent_id, 6000 value FROM dual UNION ALL
         SELECT 70 id,   60 parent_id, 7000 value FROM dual UNION ALL
         SELECT 80 id,   70 parent_id, 8000 value FROM dual),
hierarchy (id,ancestor,value,depth,path) AS (
  SELECT t.id,t.id,t.value,0,''||t.id
    FROM t
  UNION ALL
  SELECT t.id,h.ancestor,t.value,h.depth+1,h.path||'.'||t.id
    FROM t
      INNER JOIN hierarchy h
        ON t.parent_id = h.id)
SELECT LPAD(h.ancestor,p.depth*4+1,' ') AS id, t.value AS self_value, SUM(h.value) as branch_value
  FROM hierarchy h
    INNER JOIN t
      ON t.id = h.ancestor
    INNER JOIN (SELECT id,depth,path
                  FROM hierarchy
                  WHERE ancestor IN (SELECT id FROM t WHERE parent_id IS NULL)
                  ORDER BY path) p
      ON p.id = t.id
  GROUP BY h.ancestor,t.value,t.parent_id,p.path,p.depth
  ORDER BY p.path;

递归SQL：使用递归子查询因子的聚合函数

1 个答案: