我想知道是否有可能发现Oracle视图中涉及的所有表的基础主(或唯一)键列。这是一个展示我的意思的例子:
CREATE TABLE t_a (
id number(7),
primary key(id)
);
CREATE VIEW v_a AS
SELECT * FROM t_a;
因此,通过命名约定,我知道v_a.id
实际上是底层t_a
表的主键列。有没有办法使用系统视图正式发现此信息,例如SYS.ALL_CONSTRAINTS
,SYS.USER_CONSTRAINTS
等?
N.B:
答案 0 :(得分:5)
您可以通过user_dependencies视图找到该信息:
SQL> CREATE TABLE t_a
2 ( id number(7)
3 , primary key(id)
4 )
5 /
Table created.
SQL> CREATE VIEW v_a AS SELECT * FROM t_a
2 /
View created.
SQL> select c.constraint_name
2 from user_dependencies d
3 , all_constraints c
4 where d.name = 'V_A'
5 and d.referenced_type = 'TABLE'
6 and d.referenced_link_name is null
7 and d.referenced_owner = c.owner
8 and d.referenced_name = c.table_name
9 and c.constraint_type = 'P'
10 /
CONSTRAINT_NAME
------------------------------
SYS_C0051559
1 row selected.
的问候,
罗布。
编辑:对于可能的视图列名称,您可以使用此查询。请注意,您无法保证视图中存在此类列。
SQL> select c.constraint_name
2 , 'V_' || substr(c.table_name,3) || '.' || cc.column_name possible_view_column
3 from user_dependencies d
4 , all_constraints c
5 , all_cons_columns cc
6 where d.name = 'V_A'
7 and d.referenced_type = 'TABLE'
8 and d.referenced_link_name is null
9 and d.referenced_owner = c.owner
10 and d.referenced_name = c.table_name
11 and c.constraint_type = 'P'
12 and c.owner = cc.owner
13 and c.constraint_name = cc.constraint_name
14 /
CONSTRAINT_NAME POSSIBLE_VIEW_COLUMN
------------------------------ -------------------------------------------------------------
SYS_C0051561 V_A.ID
1 row selected.
答案 1 :(得分:3)
SELECT column_name FROM user_ind_columns
WHERE index_name =
(SELECT index_name FROM user_constraints
WHERE constraint_type='P' AND table_name='T_A')
ORDER BY column_position;
如果要对视图所依赖的所有表执行此操作,请将table_name上的条件更改为:
table_name IN (SELECT referenced_name FROM user_dependencies
WHERE name='view_name' AND referenced_type='TABLE')