Oracle LOCK内部机制及最佳实践系列(二)模拟RI锁定导致阻塞的场景，并分析v$lock
模拟RI锁定导致阻塞的场景，并分析v$lock相应的锁定信息，给出SQL演示。
LEO1@LEO1> create table a (id int primary key); a是主表，定义了id字段为主键
Table created.
LEO1@LEO1> create table b (id references a(id)); b是从表，id字段是引用主表的id字段
Table created.
LEO1@LEO1> insert into a values(1); 往主表a中插入一条数据但没有提交，事务没有结束会产生锁定
1 row created.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0 insert由于有从属关系因此会在2个表上都加3号共享锁
138 TM 73467 0 3 0 0
138 TX 196640 940 6 0 0
LEO1@LEO1> select object_name from dba_objects where object_id in (73465,73467); ID1就是138会话操作的对象id，我们会在主表和从表上都加上表级锁
OBJECT_NAME
--------------------------------------------------------------------------------
A 73465
B 73467
LEO1@LEO1> commit; 提交之后释放锁
Commit complete.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2; 锁会随着事务的结束而释放
no rows selected
LEO1@LEO1> select * from a; a表中有一条记录
ID
----------
1
LEO1@LEO1> select * from b;
no rows selected
LEO1@LEO1> update a set id=100 where id=1; 主表a上更新了一条记录
1 row updated.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0 update现在只对主表有锁定，从表没有锁定
138 TX 196634 941 6 0 0
LEO1@LEO1> commit; 提交之后释放锁
Commit complete.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2; 现在没有锁了
no rows selected
LEO1@LEO1> select * from a; 主表里的值已经更新了
ID
----------
100
LEO1@LEO1> delete from a;
1 row deleted.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0 delete也是只对主表有锁定，从表没有锁定
138 TX 655375 705 6 0 0
LEO1@LEO1> commit; 提交释放锁
Commit complete.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
no rows selected
LEO1@LEO1> insert into b values(2);
insert into b values(2)
*
ERROR at line 1:
ORA-02291: integrity constraint (LEO1.SYS_C0010831) violated - parent key not found 直接给从表插入记录，如果主表没有的话，会报错违反引用完整性约束，没有主表依据
LEO1@LEO1> insert into a values(2); 我们只能先给主表插入
1 row created.
LEO1@LEO1> insert into b values(2); 再给从表插入才可以，因为从表的数据必须在主表里先存在，才能正常引用
1 row created.
LEO1@LEO1> select * from a; 主表有了
ID
----------
2
LEO1@LEO1> select * from b; 从表有了
ID
----------
2
LEO1@LEO1> select * from a; 主表有3
ID
----------
2
3
LEO1@LEO1> insert into b values(3); 才能给从表插入
1 row created.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0 从表的insert也会对主从表同时加锁
138 TM 73467 0 3 0 0
138 TX 393246 939 6 0 0
LEO1@LEO1> select * from b;
ID
----------
2
3
LEO1@LEO1> commit; 释放锁
Commit complete.
LEO1@LEO1> delete from b; 删除从表
2 rows deleted.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0 从表delete也会对主从表同时加锁
138 TM 73467 0 3 0 0
138 TX 196620 944 6 0 0
LEO1@LEO1> commit; 释放锁
Commit complete.
LEO1@LEO1> select * from a; 主表有2条记录
ID
----------
2
3
LEO1@LEO1> select * from b; 从表没有记录
no rows selected
LEO1@LEO1> insert into a values(4); 向主表插入1条记录，因为没有提交所以是未决状态
1 row created.
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0 此时对主从表都加了锁定
138 TM 73467 0 3 0 0
138 TX 589834 937 6 0 0
LEO2@LEO1> insert into leo1.b values(4); 此时向从表也插入1条记录，由于从表的数据必须引用自主表，而主表数据现在是一种未决状态，所以hang住不能前进
LEO1@LEO1> select sid,type,id1,id2,lmode,request,block from v$lock where type in ('TM','TX') order by 1,2;
SID TYPE ID1 ID2 LMODE REQUEST BLOCK
---------- ---------- ---------- ---------- ---------- ---------- ----------
138 TM 73465 0 3 0 0
138 TM 73467 0 3 0 0
138 TX 589834 937 6 0 1 138会话阻塞156会话，这个6代表已经插入成功，但修改值还是未决状态
156 TM 73467 0 3 0 0
156 TM 73465 0 3 0 0
156 TX 393221 942 6 0 0 这个6代表也已经插入成功，但修改值还是未决状态
156 TX 589834 937 0 4 0 主从表插入后产生了2个TX锁，这说明这是2条不同的记录，2个独立的记录，不是争用同一条记录
小结：之所以还有一个TX锁正在申请4号锁，是因为2条记录的修改值都是未决状态违反了引用完整性约束从而产生阻塞。导致156会话hang住不能前进。
Leonarding
2012.11.28
天津&winter
分享技术~成就梦想
Blog：www.leonarding.com