大致上就是这样,SQL语句你自己可以调整看看
user_constraints
user_cons_columnsSQL> create table bb
2 (id number(8) primary key,
3 name varchar2(8));表已创建。SQL> create table cc
2 (id number(8) primary key,
3 bid number(8) references bb(id),
4 name varchar(8));表已创建。SQL> select
2 substrb(a.table_name,1,10) table_A,
3 a.constraint_type,
4 substrb(b.column_name,1,20) col
5 from user_constraints a,user_cons_columns b
6 where a.table_name='CC'
7 and a.constraint_name=b.constraint_name;TABLE_A C COL
---------- - --------------------
CC P ID
CC R BIDSQL> select
2 substrb(a.table_name,1,10) table_A,
3 a.constraint_type,
4 substrb(b.table_name,1,10) table_B,
5 substrb(b.column_name,1,20) table_B_col
6 from user_constraints a,user_cons_columns b
7 where a.table_name='CC'
8 and a.r_constraint_name=b.constraint_name;TABLE_A C TABLE_B TABLE_B_COL
---------- - ---------- --------------------
CC R BB ID
user_constraints
user_cons_columnsSQL> create table bb
2 (id number(8) primary key,
3 name varchar2(8));表已创建。SQL> create table cc
2 (id number(8) primary key,
3 bid number(8) references bb(id),
4 name varchar(8));表已创建。SQL> select
2 substrb(a.table_name,1,10) table_A,
3 a.constraint_type,
4 substrb(b.column_name,1,20) col
5 from user_constraints a,user_cons_columns b
6 where a.table_name='CC'
7 and a.constraint_name=b.constraint_name;TABLE_A C COL
---------- - --------------------
CC P ID
CC R BIDSQL> select
2 substrb(a.table_name,1,10) table_A,
3 a.constraint_type,
4 substrb(b.table_name,1,10) table_B,
5 substrb(b.column_name,1,20) table_B_col
6 from user_constraints a,user_cons_columns b
7 where a.table_name='CC'
8 and a.r_constraint_name=b.constraint_name;TABLE_A C TABLE_B TABLE_B_COL
---------- - ---------- --------------------
CC R BB ID
解决方案 »
免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货