postgresql – postgres中的简单更新查询死锁

我正在使用postgres 9.1,并在一个简单的更新方法的过度执行下获得死锁异常.

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

模式是：

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,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 );

我的猜测是,问题的根源是表中的一个循环外键引用.
TABLE vm_action_info
==> FOREIGN KEY(last_completed_vm_task_id)参考vm_task(id)
TABLE vm_task
==> FOREIGN KEY(vm_action_info_id)参考vm_action_info(id)
交易包括两个步骤：

add a new entry to task table

updates corresponding entry in vm_action_info the vm_task table.

当两个事务同时在vm_action_info表中更新相同的记录时,这将完成死锁.
看简单的测试用例：

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

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

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

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,我们在vm_action_info中向vm_task添加了一个引用id = 2的记录

session1=> begin;
BEGIN
session1=> insert into vm_task values( 100,2 );
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( 200,2 );
INSERT 0 1
session2=>

那么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;

但是这个命令挂起来正在等待锁…..
那么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！
这是因为由于外键引用,vm_task中的两个INSERT都会在vm_action_info表中的行id = 2上放置一个共享锁.然后,第一个更新尝试在此行上放置一个写入锁,因为该行被另一个(第二个)事务锁定.然后第二个更新尝试在写入模式下锁定相同的记录,但是由第一个事务锁定在共享模式.这会造成僵局.
我认为,如果您在vm_action_info中写入一个写入锁,则可以避免这种情况,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( 100,2 );
 update vm_action_info set last_on_demand_task_id=100,version=version+1 where id=2;
 commit;

postgresql – postgres中的简单更新查询死锁

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