使用connect by子句的递归插入

Big*_*oss 5 sql oracle oracle11g

我在表格中有以下方式的分层数据(右),它创建了左侧所示的层次结构.表格保存在oracle 11g中.

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

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insert语句,最好是单个insert语句,它将根据子项的值之和在数据表中插入行.请注意,我们只需要计算数据表中不存在值的父项的总和.

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

Vin*_*rat 7

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

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
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插入看起来像:

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)));
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如果叶节点的高度未知/无界,则会变得更多毛.由于ROLLUP需要确切地知道要考虑多少列,因此上述方法不起作用.

在这种情况下,您可以在自联接中使用树结构:

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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