使用connect by子句进行递归插入 [英] Recursive Insert using connect by clause

问题描述

我以下列方式在表中具有层次结构数据(右)，这创建了层次结构，如左图所示.表保存在oracle 11g中.

I have hierarchical data (right) in table in following manner which creates Hierarchy as shown in left. Tables are kept in oracle 11g.

TREE Hierarchy          Tree Table  
--------------          Element Parent
                        ------  ------
P0                      P0  
    P1                  P1      P0
        P11             P2      P0
            C111        P11     P1
            C112        P12     P1
        P12             P21     P2
            C121        P22     P2
            C122        C111    P11
    P2                  C112    P11
        P21             C121    P12
            C211        C122    P12
            C212        C211    P21
        P22             C212    P21
            C221        C221    P22
            C222        C222    P22

我的数据表的值如下.它包含所有叶节点的值.
数据表

My data table has values as follows. It contains values for all leaf nodes.
Data Table

Element Value  
C111    3  
C112    3  
C121    3  
C122    3  
C211    3  
C212    3  
C221    3  
C222    3  
P11     6  

我需要生成插入语句， 最好是单个插入语句 ，它将根据子项的值之和在数据表中插入行. 请注意，我们仅需要为数据表中未包含值的那些父母计算总和.

I need to generate insert statement, preferably single insert statement which will insert rows in data table based on sum of values of the children. Please note we need to calculate sum for only those parents whose value is not present in data table.

数据表(插入后预期)

Element Value
C111    3
C112    3
C121    3
C122    3
C211    3
C212    3
C221    3
C222    3
P11     6

-- Rows to insert
P12     6
P21     6
P22     6
P1      12
P2      12
P0      24

推荐答案

如果所有叶节点的高度都相同(此处为lvl = 4)，则可以使用ROLLUP编写一个简单的CONNECT BY查询:

If all leaf nodes are at the same height (here lvl=4), you can write a simple CONNECT BY query with a ROLLUP:

SQL> SELECT lvl0,
  2         regexp_substr(path, '[^/]+', 1, 2) lvl1,
  3         regexp_substr(path, '[^/]+', 1, 3) lvl2,
  4         SUM(VALUE) sum_value
  5    FROM (SELECT sys_connect_by_path(t.element, '/') path,
  6                 connect_by_root(t.element) lvl0,
  7                 t.element, d.VALUE, LEVEL lvl
  8             FROM tree t
  9             LEFT JOIN DATA d ON d.element = t.element
 10            START WITH t.PARENT IS NULL
 11           CONNECT BY t.PARENT = PRIOR t.element)
 12   WHERE VALUE IS NOT NULL
 13     AND lvl = 4
 14   GROUP BY lvl0, ROLLUP(regexp_substr(path, '[^/]+', 1, 2),
 15                         regexp_substr(path, '[^/]+', 1, 3));

LVL0 LVL1  LVL2   SUM_VALUE
---- ----- ----- ----------
P0   P1    P11            6
P0   P1    P12            6
P0   P1                  12
P0   P2    P21            6
P0   P2    P22            6
P0   P2                  12
P0                       24

插入内容如下:

INSERT INTO data (element, value) 
(SELECT coalesce(lvl2, lvl1, lvl0), sum_value
   FROM <query> d_out
  WHERE NOT EXISTS (SELECT NULL
                      FROM data d_in
                     WHERE d_in.element = coalesce(lvl2, lvl1, lvl0)));

如果叶节点的高度未知/无界，则将变得更加毛茸茸.由于ROLLUP需要确切知道要考虑的列数，因此上述方法无效.

If the height of the leaf nodes is unknown/unbounded this gets more hairy. The above approach wouldn't work since ROLLUP needs to know exactly how many columns are to be considered.

在这种情况下，您可以在自连接中使用树结构:

In that case, you could use the tree structure in a self-join :

SQL> WITH HIERARCHY AS (
  2     SELECT t.element, path, VALUE
  3       FROM (SELECT sys_connect_by_path(t.element, '/') path,
  4                    connect_by_isleaf is_leaf, ELEMENT
  5                FROM tree t
  6               START WITH t.PARENT IS NULL
  7              CONNECT BY t.PARENT = PRIOR t.element) t
  8       LEFT JOIN DATA d ON d.element = t.element
  9                       AND t.is_leaf = 1
 10  )
 11  SELECT h.element, SUM(elements.value)
 12    FROM HIERARCHY h
 13    JOIN HIERARCHY elements ON elements.path LIKE h.path||'/%'
 14   WHERE h.VALUE IS NULL
 15   GROUP BY h.element
 16   ORDER BY 1;

ELEMENT SUM(ELEMENTS.VALUE)
------- -------------------
P0                       24
P1                       12
P11                       6
P12                       6
P2                       12
P21                       6
P22                       6

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