4

我有一个想法,我可以编写一个查询来根据外键查找根表的所有后代表。

查询如下所示:

select level, lpad(' ', 2 * (level - 1)) || uc.table_name as "TABLE", uc.constraint_name, uc.r_constraint_name
from all_constraints uc
where uc.constraint_type in ('R', 'P')
start with uc.table_name = 'ROOT_TAB'
connect by nocycle prior uc.constraint_name = uc.r_constraint_name
order by level asc;

我得到的结果如下所示:

        1 ROOT_TAB XPKROOTTAB  
        1 ROOT_TAB R_20 XPKPART_TAB
        2 CHILD_TAB_1 R_40 XPKROOTTAB
        2 CHILD_TAB_2 R_115 XPKROOTTAB
        2 CHILD_TAB_3 R_50 XPKROOTTAB

此结果是 的所有子表ROOT_TAB,但查询不会递归到CHILD_TAB_1CHILD_TAB_2或的子表CHILD_TAB_3

递归查询对我来说是新的,所以我猜我在connect by子句中遗漏了一些东西,但我在这里画了一个空白。实际上是否有可能ROOT_TAB在单个查询中获得完整的层次结构,或者我最好将查询包装在递归过程中?

4

3 回答 3

8

你想要这样的东西:

select t.table_name, level,lpad(' ', 2 * (level - 1))||t.table_name 
from user_tables t
join user_constraints c1 
    on (t.table_name = c1.table_name 
    and c1.constraint_type in ('U', 'P'))
left join user_constraints c2 
    on (t.table_name = c2.table_name 
    and c2.constraint_type='R')
start with t.table_name = 'ROOT_TAB'
connect by prior c1.constraint_name = c2.r_constraint_name

原始查询的问题是子表的 uc.constraint_name 是外键的名称。这对于将第一个孩子连接到根表来说很好,但这不是将第二级的孩子连接到第一级所需要的。这就是为什么你需要加入两次约束——一次获取表的主键,一次获取外键。

顺便说一句,如果您要查询 all_* 视图而不是 user_* 视图,您通常希望在 table_name AND owner 上加入它们,而不仅仅是 table_name。如果多个模式具有相同名称的表,则仅加入 table_name 将给出不正确的结果。

于 2010-02-04T22:36:50.860 回答
3

对于具有多个模式和多个根表的情况,请尝试以下操作:

WITH constraining_tables AS (SELECT owner, constraint_name, table_name
                               FROM all_constraints
                              WHERE owner LIKE 'ZZZ%' AND constraint_type IN ('U', 'P')),
     constrained_tables AS (SELECT owner, constraint_name, table_name, r_owner, r_constraint_name
                              FROM all_constraints
                             WHERE owner LIKE 'ZZZ%' AND constraint_type = 'R'),
     root_tables AS (SELECT owner, table_name FROM constraining_tables
                     MINUS
                     SELECT owner, table_name FROM constrained_tables)
    SELECT c1.owner || '.' || c1.table_name, LEVEL, LPAD (' ', 2 * (LEVEL - 1)) || c1.owner || '.' || c1.table_name
      FROM    constraining_tables c1
           LEFT JOIN
              constrained_tables c2
           ON c1.owner = c2.owner AND c1.table_name = c2.table_name
START WITH c1.owner || '.' || c1.table_name IN (SELECT owner || '.' || table_name FROM root_tables)
CONNECT BY PRIOR c1.constraint_name = c2.r_constraint_name
于 2012-06-29T18:41:52.043 回答
0

经过深入调查,我制作了自己的版本来处理所有表并检索表在层次结构中的最大级别(它读取所有模式,还考虑到没有父子关系的表,与 root 一起处于级别 1那些)。如果您有访问权限,请使用 dba_ 表而不是 all_ 表。

      WITH hier AS (
                         SELECT child_table owner_table_name
                              , LEVEL lvl
                              , LPAD (' ', 4 * (LEVEL - 1)) || child_table indented_child_table
                              , sys_connect_by_path( child_table, '|' )  tree
                           FROM (
/*----------------------------------------------------------------------*/
/* Retrieve all tables. Set them as the Child column, and set their     */
/* Parent Column to NULL. This is the root list (first iteration)       */
/*----------------------------------------------------------------------*/
                                  SELECT NULL                              parent_table
                                       , a.owner || '.' || a.table_name    child_table
                                    FROM all_tables a
                                   UNION
/*----------------------------------------------------------------------*/
/* List of all possible Parent-Child relations. This table is used as   */
/* a link list, to link the current iteration with the next one, from   */
/* root to last child (last child is what we are interested to find).   */
/*----------------------------------------------------------------------*/
                                  SELECT p.owner   || '.' || p.table_name            parent_table
                                       , c.owner   || '.' || c.table_name            child_table
                                    FROM all_constraints p, all_constraints c
                                   WHERE p.owner || '.' || p.constraint_name = c.r_owner || '.' || c.r_constraint_name
                                     AND (p.constraint_type = 'P' OR p.constraint_type = 'U')
                                     AND c.constraint_type = 'R'
                                )
                     START WITH parent_table IS NULL
/*----------------------------------------------------------------------*/
/* NOCYCLE prevents infinite loops (i.e. self referencing table constr) */
/*----------------------------------------------------------------------*/
                     CONNECT BY NOCYCLE PRIOR child_table = parent_table
                   )
                     SELECT *
                       FROM hier
                      WHERE (owner_table_name, lvl) IN (   SELECT owner_table_name
                                                                , MAX(lvl)
                                                             FROM hier
                                                         GROUP BY owner_table_name
                                                       );

编辑:在查找无限循环时,此查询存在“某种”问题。

如果我们有这棵树:

b --> c --> d
b <-- c

它将 lvl 2 分配给 c 作为: 将 lvl 2 分配b --> c 给 b 作为:c --> b

对于 d,它会检测到b --> c --> d ,所以它会分配 lvl 3

如您所见,问题出在循环内部,来自外部的值将始终具有其最大正确 lvl

于 2017-04-29T20:18:42.987 回答