我正在尝试理解MySQL在处理相同表的并发客户端时发现的死锁。以下是“SHOW InnoDB STATUS”命令的有趣部分：
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LATEST DETECTED DEADLOCK
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120704 16:17:51
*** (1) TRANSACTION:
TRANSACTION 0 3547576, ACTIVE 0 sec, process no 10886, OS thread id 140547111458560 inserting
mysql tables in use 1, locked 1
LOCK WAIT 2 lock struct(s), heap size 368, 1 row lock(s), undo log entries 1
MySQL thread id 41941, query id 1725666 localhost testsuite update
insert into `INode` (`status`,`_type`,`group`,`ctime`,`parent`,`shared`,`basename`,`_rowid`,`displayname`,`user`,`content_type`,`mtime`,`position`,`atime`,`size`) values ('Published','Group','12','2012-07-04 16:17:48.996869','2',null,'1','12','1','3','application/x-empty','2012-07-04 16:17:48.996896','1','2012-07-04 16:17:48.996914',null)
*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 0 page no 5554 n bits 80 index `INodeparent_basename` of table `testsuite`.`INode` trx id 0 3547576 lock mode S waiting
Record lock, heap no 12 PHYSICAL RECORD: n_fields 3; compact format; info bits 32
0: len 4; hex 80000002; asc ;; 1: len 1; hex 31; asc 1;; 2: len 4; hex 8000000b; asc ;;
*** (2) TRANSACTION:
TRANSACTION 0 3547575, ACTIVE 0 sec, process no 10886, OS thread id 140547107845888 inserting, thread declared inside InnoDB 493
mysql tables in use 1, locked 1
13 lock struct(s), heap size 3024, 17 row lock(s), undo log entries 21
MySQL thread id 41940, query id 1725808 localhost testsuite update
replace into `INode` (`status`,`_type`,`position`,`group`,`ctime`,`parent`,`basename`,`_rowid`,`displayname`,`user`,`content_type`,`mtime`,`shared`,`atime`,`size`) values ('Published','Group','0','2','2012-07-04 16:17:49','1','groups','2','admin','3','application/x-empty','2012-07-04 16:17:49',null,'2012-07-04 16:17:49',null),('Published','Group','1','11','2012-07-04 16:17:51.064074','2','1','11','1','3','inode/directory','2012-07-04 16:17:51.064074',null,'2012-07-04 16:17:51.064074',null)
*** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 0 page no 5554 n bits 80 index `INodeparent_basename` of table `testsuite`.`INode` trx id 0 3547575 lock_mode X locks rec but not gap
Record lock, heap no 3 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
0: len 4; hex 80000001; asc ;; 1: len 6; hex 67726f757073; asc groups;; 2: len 4; hex 80000002; asc ;;
Record lock, heap no 12 PHYSICAL RECORD: n_fields 3; compact format; info bits 32
0: len 4; hex 80000002; asc ;; 1: len 1; hex 31; asc 1;; 2: len 4; hex 8000000b; asc ;;
*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 0 page no 5554 n bits 80 index `INodeparent_basename` of table `testsuite`.`INode` trx id 0 3547575 lock_mode X waiting
Record lock, heap no 12 PHYSICAL RECORD: n_fields 3; compact format; info bits 32
0: len 4; hex 80000002; asc ;; 1: len 1; hex 31; asc 1;; 2: len 4; hex 8000000b; asc ;;
*** WE ROLL BACK TRANSACTION (1)
testsuite的索引INodeparent_basename上的共享锁。现在，事务(2)正在该索引上拥有独占锁。但奇怪的是，事务(2)还在等待给定表上的X锁。如果事务(2)已经拥有该锁，为什么它还要等待呢？（总的来说，我没有找到任何文件确切地说明如何阅读状态命令的输出 - 如果有人能指出给我，那将很有趣）