带有引发触发器的插入的 MySQL 死锁

Question

我发现发生了一些罕见的死锁错误。我知道当两个查询作业相互依赖时会发生死锁，因此 MySQL 会回滚其中一个。但在我的情况下，MySQL 处于自动提交模式，我正在插入一条触发触发器的新记录。所以我不明白它产生死锁情况的原因。

这是我的表架构:

---- 用户表----

CREATE TABLE `users` (
`insta_id` bigint(20) unsigned NOT NULL,
`name` varchar(50) NOT NULL,
`password` varchar(60) NOT NULL,
`gem` int(10) unsigned DEFAULT '20',
`coin` int(10) unsigned DEFAULT '20',
 PRIMARY KEY (`insta_id`),
 UNIQUE KEY `insta_id` (`insta_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1

---like_requests表---

CREATE TABLE `like_requests` (
`req_id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`insta_id` bigint(20) unsigned NOT NULL,
`media_id` varchar(50) NOT NULL,
`remaining_like` int(10) unsigned NOT NULL,
`active` tinyint(1) NOT NULL DEFAULT '1',
`count` int(10) unsigned NOT NULL,
PRIMARY KEY (`req_id`),
KEY `insta_id` (`insta_id`),
KEY `media_id` (`media_id`),
CONSTRAINT `like_requests_ibfk_1` FOREIGN KEY (`insta_id`) REFERENCES `users`(`insta_id`)
) ENGINE=InnoDB AUTO_INCREMENT=103902 DEFAULT CHARSET=latin1

---喜欢表格---

CREATE TABLE `likes` (
 `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
 `insta_id` bigint(20) unsigned NOT NULL,
 `media_id` varchar(50) NOT NULL,
 `req_id` bigint(20) unsigned DEFAULT NULL,
 `date` timestamp NULL DEFAULT CURRENT_TIMESTAMP,
 PRIMARY KEY (`id`),
 UNIQUE KEY `id` (`id`),
 KEY `req_id` (`req_id`),
 KEY `insta_id` (`insta_id`),
 KEY `media_id` (`media_id`),
 CONSTRAINT `likes_ibfk_1` FOREIGN KEY (`req_id`) REFERENCES  `like_requests`(`req_id`),
 CONSTRAINT `likes_ibfk_2` FOREIGN KEY (`insta_id`) REFERENCES `users`(`insta_id`)
) ENGINE=InnoDB AUTO_INCREMENT=1704209 DEFAULT CHARSET=latin1

我在 likes 表上有一个触发器，定义如下:

CREATE TRIGGER `after_insert_likes` AFTER INSERT ON `likes`
    FOR EACH ROW BEGIN
        UPDATE users SET users.coin=users.coin+1
            WHERE users.insta_id = NEW.insta_id  LIMIT 1;
        IF NEW.req_id IS NOT NULL THEN
            UPDATE like_requests
                SET  like_requests.remaining_like = like_requests.remaining_like-1
                WHERE like_requests.req_id = NEW.req_id
                  AND like_requests.remaining_like > 0
                LIMIT 1;
        END IF;
    END

做一些简单的插入:

    $sql = "INSERT INTO likes (insta_id,media_id,req_id) VALUES (?,?,?);";
    $pdo = $this->db;

    $statement = $pdo->prepare($sql);
    $statement->bindValue(1,$data['id'],PDO::PARAM_INT);
    $statement->bindValue(2,$data['media_id']);
    $statement->bindValue(3,$data['req_id'],PDO::PARAM_INT);

    try
    {
        $statement->execute();
        return GetOkResponseWithMessage($response,"Like was submitted");
    }
    catch (PDOException $exc)
    {
        return GetErrorResponseWithMessage($response,$exc->getMessage(),500);
    }

我得到以下死锁错误日志:

*** (1) TRANSACTION:
TRANSACTION 29031910, ACTIVE 1 sec starting index read
mysql tables in use 4, locked 4
LOCK WAIT 7 lock struct(s), heap size 1184, 3 row lock(s), undo log entries 1
MySQL thread id 264238, OS thread handle 0x7f6522c6eb00, query id 753506        localhost xxxx updating
UPDATE users SET users.coin=users.coin+1 WHERE users.insta_id=NEW.insta_id   LIMIT 1
*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 14 page no 1560 n bits 128 index `PRIMARY` of table  `insta_star`.`users` trx table locks 4 total table locks 4  trx id 29031910  lock_mode X locks rec but not gap waiting lock hold time 0 wait time before grant  0
*** (2) TRANSACTION:
TRANSACTION 29031909, ACTIVE 1 sec starting index read
mysql tables in use 4, locked 4
7 lock struct(s), heap size 1184, 3 row lock(s), undo log entries 1
MySQL thread id 264237, OS thread handle 0x7f65209f8b00, query id 753507 localhost xxxx updating
UPDATE users SET users.coin=users.coin+1 WHERE users.insta_id=NEW.insta_id   LIMIT 1
*** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 14 page no 1560 n bits 128 index `PRIMARY` of table  `insta_star`.`users` trx table locks 4 total table locks 4  trx id 29031909 lock   mode S locks rec but not gap lock hold time 0 wait time before grant 0
*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 14 page no 1560 n bits 128 index `PRIMARY` of table    `insta_star`.`users` trx table locks 4 total table locks 4  trx id 29031909   lock_mode X locks rec but not gap waiting lock hold time 0 wait time before grant   0
*** WE ROLL BACK TRANSACTION (2)

这不应该以锁等待而不是死锁结束吗？

如何在不重启交易的情况下解决这个问题？

Answer 1

线程 2 持有用户表中行的共享锁。

然后线程1尝试获取同一行的排他锁，进入锁等待。

但是线程 1 不会有机会超时，因为线程 2 然后试图将他的锁升级为独占...但是要做到这一点，他必须等待处于锁等待状态的线程 1，但是它正在等待线程 2。

他们互相阻挡。

这是一个僵局。

服务器选择要终止的事务，这样它们就不会不必要地相互阻塞。

死锁检测允许一个线程以另一个线程为代价立即成功。否则他们都会被困在等待中，直到其中一个因等待时间过长而死亡。

---

您处于自动提交模式，当然，这并不意味着您不在事务中。 InnoDB 的每个查询仍然在事务中处理，但使用自动提交时，事务在查询开始执行时隐式启动，并在查询成功时隐式提交。