我们可以使用基于函数的索引来做到这一点。如您所知,以下使用NVL2()
它,如果表达式不为空,则返回一个值,如果为空,则返回不同的值。你可以CASE()
改用。
SQL> create table blah (name varchar2(10), email varchar2(20))
2 /
Table created.
SQL> create unique index blah_uidx on blah
2 (nvl2(email, name, null), nvl2(name, email, null))
3 /
Index created.
SQL> insert into blah values ('APC', null)
2 /
1 row created.
SQL> insert into blah values ('APC', null)
2 /
1 row created.
SQL> insert into blah values (null, 'apc@example.com')
2 /
1 row created.
SQL> insert into blah values (null, 'apc@example.com')
2 /
1 row created.
SQL> insert into blah values ('APC', 'apc@example.com')
2 /
1 row created.
SQL> insert into blah values ('APC', 'apc@example.com')
2 /
insert into blah values ('APC', 'apc@example.com')
*
ERROR at line 1:
ORA-00001: unique constraint (APC.BLAH_UIDX) violated
SQL>
编辑
因为在您的方案名称中将始终填充您只需要这样的索引:
SQL> create unique index blah_uidx on blah
2 (nvl2(email, name, null), email)
3 /
Index created.
SQL>