未加索引的外键导致的死锁模拟实验以及问题
EODA@PROD1> create table pk_t (x int primary key); --建立主角
Table created.
EODA@PROD1> insert into pk_t select rownum from dual connect by level <= 40;
40 rows created.
EODA@PROD1> commit;
Commit complete.
EODA@PROD1> create table fk_t (x int , y int); --建立副表
Table created.
EODA@PROD1> insert into fk_t select rownum, rownum/2 from dual connect by level <= 30;
30 rows created.
EODA@PROD1> commit;
Commit complete.
EODA@PROD1> alter table fk_t add constraint fk_fkt foreign key (x) references pk_t(x); --创建外键约束
Table altered.
EODA@PROD1> select table_name, constraint_name,
2 cname1 || nvl2(cname2,','||cname2,null) ||
3 nvl2(cname3,','||cname3,null) || nvl2(cname4,','||cname4,null) ||
4 nvl2(cname5,','||cname5,null) || nvl2(cname6,','||cname6,null) ||
5 nvl2(cname7,','||cname7,null) || nvl2(cname8,','||cname8,null)
6 columns
7 from ( select b.table_name,
8 b.constraint_name,
9 max(decode( position, 1, column_name, null )) cname1,
10 max(decode( position, 2, column_name, null )) cname2,
11 max(decode( position, 3, column_name, null )) cname3,
12 max(decode( position, 4, column_name, null )) cname4,
13 max(decode( position, 5, column_name, null )) cname5,
14 max(decode( position, 6, column_name, null )) cname6,
15 max(decode( position, 7, column_name, null )) cname7,
16 max(decode( position, 8, column_name, null )) cname8,
17 count(*) col_cnt
18 from (select substr(table_name,1,30) table_name,
19 substr(constraint_name,1,30) constraint_name,
20 substr(column_name,1,30) column_name,
21 position
22 from user_cons_columns ) a,
23 user_constraints b
24 where a.constraint_name = b.constraint_name
25 and b.constraint_type = 'R'
26 group by b.table_name, b.constraint_name
27 ) cons
28 where col_cnt > ALL
29 ( select count(*)
30 from user_ind_columns i,
31 user_indexes ui
32 where i.table_name = cons.table_name
33 and i.column_name in (cname1, cname2, cname3, cname4,
34 cname5, cname6, cname7, cname8 )
35 and i.column_position <= cons.col_cnt
36 and ui.table_name = i.table_name
37 and ui.index_name = i.index_name
38 and ui.index_type IN ('NORMAL','NORMAL/REV')
39 group by i.index_name
40 )
41 /
--查找出未加索引的外键
TABLE_NAME CONSTRAINT_NAME COLUMNS
--------------- --------------- ------------------------------
FK_T FK_FKT X
……
--模拟死锁
EODA@PROD1> delete from fk_t where x=1; --第一个会话中执行删除，子表加上SX锁
1 row deleted.
EODA@PROD1> delete from pk_t where x=1; --第一个会话中执行删除，父表加上SX锁
1 row deleted.
EODA@PROD1> delete from fk_t where x=2; --第二个会话中执行删除
1 row deleted.
EODA@PROD1> delete from pk_t where x=2; --第二个会话中执行删除，hang住
EODA@PROD1> delete from pk_t where x=2; --再次在第一个会话中执行删除
EODA@PROD1> delete pk_t where x=2;
delete pk_t where x=2
*
ERROR at line 1:
ORA-00060: deadlock detected while waiting for resource --第二个会话中立刻报错形成死锁
--再次模拟死锁
EODA@PROD1> create index idx_fk_x on fk_t(x) nologging; --给外键增加索引
Index created.
EODA@PROD1> delete from fk_t where x=1; --第一个会话中执行删除，子表加上SX锁
1 row deleted.
EODA@PROD1> delete from pk_t where x=1; --第一个会话中执行删除，父表加上SX锁
1 row deleted.
EODA@PROD1> delete from fk_t where x=2; --第二个会话中执行删除
1 row deleted.
EODA@PROD1> delete from pk_t where x=2; --第二个会话中执行删除，均成功。
1 row deleted.--未加索引的外键会存在的问题
1.引起全表锁。
2.当使用了on delete cascade，而且没有对子表加索引，要从父表中删除多行时，每删除一行就要扫描一次子表。
3.当你从父表查询子表，你会发现没有索引的话会使查询变慢。
--什么时候不需要对外键增加索引
1.不会从父表中删除行。
2.不会去更新父表的主键。
3.不会从父表连接到子表。
--参考来源《Oracle编程艺术深入理解数据库体系结构（第三版）》