关于简单更新查询的postgres中的死锁

时间：2013-08-18 09:01:37

标签： postgresql postgresql-9.1

我正在使用postgres 9.1并在过度执行简单更新方法时遇到死锁异常。

根据日志，由于同时执行两个相同的更新而发生死锁。

  

更新public.vm_action_info设置last_on_demand_task_id = $ 1，版本=版本+ 1

两个相同的简单更新如何相互死锁？

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 );

2 个答案:

答案 0 :(得分：18)

我的猜测是问题的根源是表格中的循环外键引用。

表vm_action_info
==＆GT; FOREIGN KEY（last_completed_vm_task_id）REFERENCES vm_task（id）

表vm_task
==＆GT; FOREIGN KEY（vm_action_info_id）参考vm_action_info（id）

该交易包括两个步骤：

  
      
  1. 向任务表添加新条目
    2.   
  2. 更新vm_action_info vm_task表中的相应条目。
    3.   

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。

看一下简单的测试用例：

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的同一时间，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;

但此命令挂起并等待锁定.....

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;
BŁĄD:  wykryto zakleszczenie
SZCZEGÓŁY:  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.
PODPOWIEDŹ:  Przejrzyj dziennik serwera by znaleźć szczegóły 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。这会造成僵局。

我认为如果你在vm_action_info中记录一个写锁，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;

答案 1 :(得分：2)

可能只是您的系统异常忙碌。你说你只是通过“过度执行”查询来看到这一点。

看来情况如下：

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

所以 - 我们似乎有四个交易同时更新了这一行。事务4228275尚未提交或回滚，并且正在阻止其他事务。其中两个已等待deadlock_timeout秒，否则我们看不到超时。 Timout到期，死锁检测器看一看，看到一堆交织在一起的交易并取消其中一个。可能不是严格意义上的僵局，但我不确定探测器是否足够聪明，可以解决这个问题。

尝试以下方法之一：

  1. 降低更新率
  2. 获取更快的服务器
  3. 增加deadlock_timeout

    4. 可能＃3是最简单的:-)可能也想设置log_lock_waits所以你可以看看你的系统是否/何时处于这种压力之下。

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