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我正在尝试解决涉及我们其中一张繁忙表上的死锁的错误。我已经阅读了这个关于死锁的 SO 问题，虽然它是有道理的，但在我的情况下，查询顺序似乎并不是原因。

这是的缩写输出SHOW ENGINE INNODB STATUS;：

*** (1) TRANSACTION:
TRANSACTION 1 2611184895, ACTIVE 0 sec, process no 17501, OS thread id 140516779579136 starting index read
mysql tables in use 1, locked 1
LOCK WAIT 2 lock struct(s), heap size 368, 1 row lock(s)
MySQL thread id 211935717, query id 3146186174 [SERVER A] Searching rows for update

UPDATE images_unread_comments
    SET unread = 0
    WHERE user_id = 1 AND comment_id IN(1,2,3) AND unread = 1

*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 0 page no 404976 n bits 632 index `users_unread_comments` of table images_unread_comments trx id 1 2611184895 lock_mode X waiting
Record lock, heap no 558 PHYSICAL RECORD: n_fields 3; compact format; info bits 32
 0: len 4; hex 0001461a; asc   F ;; 1: len 1; hex 01; asc  ;; 2: len 6; hex 00000e67d888; asc    g  ;;

*** (2) TRANSACTION:
TRANSACTION 1 2611184892, ACTIVE 0 sec, process no 17501, OS thread id 140516774520576 updating or deleting, thread declared inside InnoDB 494
mysql tables in use 1, locked 1
6 lock struct(s), heap size 1216, 11 row lock(s), undo log entries 1
MySQL thread id 211935715, query id 3146186169 [SERVER B] Updating

UPDATE images_unread_comments
    SET unread = 0
    WHERE user_id = 1 AND comment_id IN(1,2,3) AND unread = 1
    *** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 0 page no 404976 n bits 632 index users_unread_comments of table images_unread_comments trx id 1 2611184892 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;;

Record lock, heap no 555 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
 0: len 4; hex 0001461a; asc   F ;; 1: len 1; hex 01; asc  ;; 2: len 6; hex 00000e67daf0; asc    g  ;;

Record lock, heap no 556 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
 0: len 4; hex 0001461a; asc   F ;; 1: len 1; hex 01; asc  ;; 2: len 6; hex 00000e67dadb; asc    g  ;;

Record lock, heap no 557 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
 0: len 4; hex 0001461a; asc   F ;; 1: len 1; hex 01; asc  ;; 2: len 6; hex 00000e67d940; asc    g @;;

Record lock, heap no 558 PHYSICAL RECORD: n_fields 3; compact format; info bits 32
 0: len 4; hex 0001461a; asc   F ;; 1: len 1; hex 01; asc  ;; 2: len 6; hex 00000e67d888; asc    g  ;;

*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 0 page no 404976 n bits 632 index users_unread_comments of table images_unread_comments trx id 1 2611184892 lock_mode X locks gap before rec insert intention waiting
Record lock, heap no 558 PHYSICAL RECORD: n_fields 3; compact format; info bits 32
 0: len 4; hex 0001461a; asc   F ;; 1: len 1; hex 01; asc  ;; 2: len 6; hex 00000e67d888; asc    g  ;;

*** WE ROLL BACK TRANSACTION (1)

我注意到这两个 SQL 语句是相同的；但是，一个在服务器 A 上执行，另一个在服务器 B 上执行。不管发生这种情况的原因 - 如果两个查询以相同的顺序锁定相同的键，为什么会造成死锁？还是我首先误解了死锁的原因？

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1 回答 1

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似乎事务 1 执行了另一个操作（插入？），其中它锁定了索引中的一个间隙。然后它等待事务 2 执行更新，因为事务 2 已锁定 ID 为 1 的记录。但是事务 2 无法继续，因为事务 1 持有索引上的锁定。如果可以通过该操作隔离事务中使用的所有 SQL 语句，我们可以看到死锁的确切原因

于 2013-07-06T18:25:43.727 回答