对于这两个插入查询，我有一个死锁：

insert into PlayerClub (modifiedBy, timeCreated, currentClubId, endingLevelPosition, nextClubId, account_id) values (0, '2014-12-23 15:47:11.596', 180, 4, 181, 561)

insert into PlayerClub (modifiedBy, timeCreated, currentClubId, endingLevelPosition, nextClubId, account_id) values (0, '2014-12-23 15:47:11.611', 180, 4, 181, 563)

这是InnoDB状态：

------------------------
LATEST DETECTED DEADLOCK
------------------------
2014-12-23 15:47:11 1f4c
*** (1) TRANSACTION:
TRANSACTION 19896526, ACTIVE 0 sec inserting
mysql tables in use 1, locked 1
LOCK WAIT 5 lock struct(s), heap size 1248, 3 row lock(s), undo log entries 1
MySQL thread id 17988, OS thread handle 0x17bc, query id 5701353 localhost 127.0.0.1 root update
insert into PlayerClub (modifiedBy, timeCreated, currentClubId, endingLevelPosition,  nextClubId, account_id) values (0, '2014-12-23 15:47:11.596', 180, 4, 181, 561)
*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 49735 page no 4 n bits 72 index `UK_cagoa3q409gsukj51ltiokjoh` of   table `db`.`playerclub` trx id 19896526 lock_mode X insert intention waiting
Record lock, heap no 1 PHYSICAL RECORD: n_fields 1; compact format; info bits 0
 0: len 8; hex 73757072656d756d; asc supremum;;

*** (2) TRANSACTION:
TRANSACTION 19896542, ACTIVE 0 sec inserting, thread declared inside InnoDB 5000
mysql tables in use 1, locked 1
5 lock struct(s), heap size 1248, 3 row lock(s), undo log entries 1
MySQL thread id 17979, OS thread handle 0x1f4c, query id 5701360 localhost 127.0.0.1    root update
insert into PlayerClub (modifiedBy, timeCreated, currentClubId, endingLevelPosition,   nextClubId, account_id) values (0, '2014-12-23 15:47:11.611', 180, 4, 181, 563)
*** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 49735 page no 4 n bits 72 index `UK_cagoa3q409gsukj51ltiokjoh` of   table `db`.`playerclub` trx id 19896542 lock_mode X
Record lock, heap no 1 PHYSICAL RECORD: n_fields 1; compact format; info bits 0
 0: len 8; hex 73757072656d756d; asc supremum;;

*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 49735 page no 4 n bits 72 index `UK_cagoa3q409gsukj51ltiokjoh` of    table `db`.`playerclub` trx id 19896542 lock_mode X insert intention waiting
Record lock, heap no 1 PHYSICAL RECORD: n_fields 1; compact format; info bits 0
 0: len 8; hex 73757072656d756d; asc supremum;;

*** WE ROLL BACK TRANSACTION (2)

该表上唯一的foreign键是“ account_id”。

有任何想法吗？

编辑：这是我的PlayerClub信息：

CREATE TABLE `PlayerClub` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `modifiedBy` bigint(20) DEFAULT NULL,
  `timeCreated` datetime NOT NULL,
  `account_id` bigint(20) DEFAULT NULL,
  `currentClubId` bigint(20) DEFAULT NULL,
  `endingLevelPosition` int(11) NOT NULL,
  `nextClubId` bigint(20) DEFAULT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `UK_cagoa3q409gsukj51ltiokjoh` (`account_id`),
  KEY `FK_cagoa3q409gsukj51ltiokjoh` (`account_id`),
  CONSTRAINT `FK_cagoa3q409gsukj51ltiokjoh` FOREIGN KEY (`account_id`) REFERENCES   `PlayerAccount` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=latin1

这是事实

这是两个INSERT

insert into PlayerClub (modifiedBy, timeCreated, currentClubId, endingLevelPosition, nextClubId, account_id) values (0, '2014-12-23 15:47:11.596', 180, 4, 181, 561)
insert into PlayerClub (modifiedBy, timeCreated, currentClubId, endingLevelPosition, nextClubId, account_id) values (0, '2014-12-23 15:47:11.611', 180, 4, 181, 563)

这是您的两行 SHOW ENGINE INNODB STATUS\G

RECORD LOCKS space id 49735 page no 4 n bits 72 index `UK_cagoa3q409gsukj51ltiokjoh` of   table `db`.`playerclub` trx id 19896526 lock_mode X insert intention waiting
RECORD LOCKS space id 49735 page no 4 n bits 72 index `UK_cagoa3q409gsukj51ltiokjoh` of   table `db`.`playerclub` trx id 19896542 lock_mode X

观察

您用两个不同的account_ids进行INSERT：561和563。

它们是唯一的，不应有问题，对吗？错误！

由于InnoDB的聚集索引，仍然存在僵局。为什么呢

回头看看您的两个INSERT。在PRIMARY KEY对ID中没有规定。它必须是自动生成的。除主键（唯一或非唯一）以外的任何键都将附加主键。

请注意有关如何将二级索引和主键交织在一起的MySQL文档：

除聚集索引以外的所有索引都称为辅助索引。在InnoDB中，二级索引中的每个记录都包含该行的主键列以及为二级索引指定的列。InnoDB使用此主键值在聚集索引中搜索行。

如果主键较长，则辅助索引将使用更多空间，因此具有较短的主键是有利的。

尽管您要插入account_id 561和563，但在幕后您将插入561-(id)并563-(id)插入UK_cagoa3q409gsukj51ltiokjoh索引。之所以PRIMARY KEY成为瓶颈，是因为二级索引必须等待直到该id列自动生成。

建议

您有一个带有两个候选键的表

• PRIMARY KEY 上 id
• UNIQUE KEY 上 UK_cagoa3q409gsukj51ltiokjoh

由于两者都是BIGINT，您可以PlayerClub通过避免并避免这种死锁情况来提高性能并通过消除id并保持唯一性来减小表的大小UK_cagoa3q409gsukj51ltiokjoh。