关于简单更新查询的postgres中的死锁

Question

我正在使用postgres 9.1并在过度执行简单更新方法时获得死锁异常.

根据日志,由于同时执行两个相同的更新而发生死锁.

update public.vm_action_info set last_on_demand_task_id = $ 1,version = version + 1

两个相同的简单更新如何相互僵持？

我在日志中收到的错误

2013-08-18 11:00:24 IDT HINT:  See server log for query details.
2013-08-18 11:00:24 IDT STATEMENT:  update public.vm_action_info set last_on_demand_task_id=$1, version=version+1 where id=$2
2013-08-18 11:00:25 IDT ERROR:  deadlock detected
2013-08-18 11:00:25 IDT DETAIL:  Process 31533 waits for ShareLock on transaction 4228275; blocked by process 31530.
        Process 31530 waits for ExclusiveLock on tuple (0,68) of relation 70337 of database 69205; blocked by process 31533.
        Process 31533: update public.vm_action_info set last_on_demand_task_id=$1, version=version+1 where id=$2
        Process 31530: update public.vm_action_info set last_on_demand_task_id=$1, version=version+1 where id=$2
2013-08-18 11:00:25 IDT HINT:  See server log for query details.
2013-08-18 11:00:25 IDT STATEMENT:  update public.vm_action_info set last_on_demand_task_id=$1, version=version+1 where id=$2
2013-08-18 11:00:25 IDT ERROR:  deadlock detected
2013-08-18 11:00:25 IDT DETAIL:  Process 31530 waits for ExclusiveLock on tuple (0,68) of relation 70337 of database 69205; blocked by process 31876.
        Process 31876 waits for ShareLock on transaction 4228275; blocked by process 31530.
        Process 31530: update public.vm_action_info set last_on_demand_task_id=$1, version=version+1 where id=$2
        Process 31876: update public.vm_action_info set last_on_demand_task_id=$1, version=version+1 where id=$2

架构是:

CREATE TABLE vm_action_info(
  id integer NOT NULL,
  version integer NOT NULL DEFAULT 0,
  vm_info_id integer NOT NULL,
 last_exit_code integer,
  bundle_action_id integer NOT NULL,
  last_result_change_time numeric NOT NULL,
  last_completed_vm_task_id integer,
  last_on_demand_task_id bigint,
  CONSTRAINT vm_action_info_pkey PRIMARY KEY (id ),
  CONSTRAINT vm_action_info_bundle_action_id_fk FOREIGN KEY (bundle_action_id)
      REFERENCES bundle_action (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE CASCADE,
  CONSTRAINT vm_discovery_info_fk FOREIGN KEY (vm_info_id)
      REFERENCES vm_info (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE CASCADE,
  CONSTRAINT vm_task_last_on_demand_task_fk FOREIGN KEY (last_on_demand_task_id)
      REFERENCES vm_task (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION,

  CONSTRAINT vm_task_last_task_fk FOREIGN KEY (last_completed_vm_task_id)
      REFERENCES vm_task (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION
)
WITH (OIDS=FALSE);

ALTER TABLE vm_action_info
  OWNER TO vadm;

-- Index: vm_action_info_vm_info_id_index

-- DROP INDEX vm_action_info_vm_info_id_index;

CREATE INDEX vm_action_info_vm_info_id_index
  ON vm_action_info
  USING btree (vm_info_id );

CREATE TABLE vm_task
(
  id integer NOT NULL,
  version integer NOT NULL DEFAULT 0,
  vm_action_info_id integer NOT NULL,
  creation_time numeric NOT NULL DEFAULT 0,
  task_state text NOT NULL,
  triggered_by text NOT NULL,
  bundle_param_revision bigint NOT NULL DEFAULT 0,
  execution_time bigint,
  expiration_time bigint,
  username text,
  completion_time bigint,
  completion_status text,
  completion_error text,
  CONSTRAINT vm_task_pkey PRIMARY KEY (id ),
  CONSTRAINT vm_action_info_fk FOREIGN KEY (vm_action_info_id)
  REFERENCES vm_action_info (id) MATCH SIMPLE
  ON UPDATE NO ACTION ON DELETE CASCADE
)
 WITH (
OIDS=FALSE
);
ALTER TABLE vm_task
  OWNER TO vadm;

-- Index: vm_task_creation_time_index

-- DROP INDEX vm_task_creation_time_index     ;

CREATE INDEX vm_task_creation_time_index
  ON vm_task
  USING btree
 (creation_time );

Answer 1

我的猜测是问题的根源是表中的循环外键引用.

表vm_action_info
==> FOREIGN KEY(last_completed_vm_task_id)参考vm_task(id)

表vm_task
==> FOREIGN KEY(vm_action_info_id)参考vm_action_info(id)

事务包含两个步骤:

1. 向任务表添加新条目
2. 更新vm_action_info vm_task表中的相应条目.

当两个事务要同时更新vm_action_info表中的同一记录时,这将以死锁结束.

看看简单的测试用例:

CREATE TABLE vm_task
(
  id integer NOT NULL,
  version integer NOT NULL DEFAULT 0,
  vm_action_info_id integer NOT NULL,
  CONSTRAINT vm_task_pkey PRIMARY KEY (id )
)
 WITH ( OIDS=FALSE );

 insert into vm_task values 
 ( 0, 0, 0 ), ( 1, 1, 1 ), ( 2, 2, 2 );

CREATE TABLE vm_action_info(
  id integer NOT NULL,
  version integer NOT NULL DEFAULT 0,
  last_on_demand_task_id bigint,
  CONSTRAINT vm_action_info_pkey PRIMARY KEY (id )
)
WITH (OIDS=FALSE);
insert into vm_action_info values 
 ( 0, 0, 0 ), ( 1, 1, 1 ), ( 2, 2, 2 );

alter table vm_task
add  CONSTRAINT vm_action_info_fk FOREIGN KEY (vm_action_info_id)
  REFERENCES vm_action_info (id) MATCH SIMPLE
  ON UPDATE NO ACTION ON DELETE CASCADE
  ;
Alter table vm_action_info
 add CONSTRAINT vm_task_last_on_demand_task_fk FOREIGN KEY (last_on_demand_task_id)
      REFERENCES vm_task (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION
      ;

在会话1中,我们向vm_task添加一条记录,该记录在vm_action_info中引用id = 2

session1=> begin;
BEGIN
session1=> insert into vm_task values( 100, 0, 2 );
INSERT 0 1
session1=>

在会话2的同时,另一个交易开始:

session2=> begin;
BEGIN
session2=> insert into vm_task values( 200, 0, 2 );
INSERT 0 1
session2=>

然后第一个事务执行更新:

session1=> update vm_action_info set last_on_demand_task_id=100, version=version+1
session1=> where id=2;

但是这个命令挂起并等待锁.....

然后第二个会话执行更新........

session2=> update vm_action_info set last_on_demand_task_id=200, version=version+1 where id=2;
B??D:  wykryto zakleszczenie
SZCZEGÓ?Y:  Proces 9384 oczekuje na ExclusiveLock na krotka (0,5) relacji 33083 bazy danych 16393; zablokowany przez 380
8.
Proces 3808 oczekuje na ShareLock na transakcja 976; zablokowany przez 9384.
PODPOWIED?:  Przejrzyj dziennik serwera by znale?? szczegó?y zapytania.
session2=>

检测到死锁!!!

这是因为对于vm_task的INSERT都由于外键引用而在vm_action_info表中的行id = 2上放置共享锁.然后第一次更新尝试在此行上放置写锁并挂起,因为该行被另一个(第二个)事务锁定.然后第二次更新尝试在写入模式下锁定相同的记录,但第一次事务将其锁定在共享模式下.这导致了僵局.

我认为如果你在vm_action_info中的记录上放置一个写锁,这可以避免,整个事务必须包含5个步骤:

 begin;
 select * from vm_action_info where id=2 for update;
 insert into vm_task values( 100, 0, 2 );
 update vm_action_info set last_on_demand_task_id=100, 
         version=version+1 where id=2;
 commit;

Answer 2

可能只是您的系统异常繁忙。您说您只在查询“过度执行”时看到过这种情况。

看起来的情况是这样的：

pid=31530 wants to lock tuple (0,68) on rel 70337 (vm_action_info I suspect) for update
    it is waiting behind pid=31533, pid=31876
pid=31533 is waiting behind transaction 4228275
pid=31876 is waiting behind transaction 4228275

因此，我们似乎有四个事务同时更新该行。事务 4228275 尚未提交或回滚，并且阻碍了其他事务。其中两个一直在等待deadlock_timeout秒，否则我们不会看到超时。超时到期，死锁检测器查看，发现一堆相互交织的事务并取消其中一个。严格来说可能不是僵局，但我不确定探测器是否足够聪明来解决这个问题。

尝试以下之一：

1. 降低更新率
2. 获得更快的服务器
3. 增加 deadlock_timeout

可能 #3 是最简单的:-) 可能也想设置 log_lock_waits 以便您可以查看您的系统是否/何时处于这种压力之下。