我發現了一些罕見的死鎖錯誤發生。我知道當兩個查詢工作取決於對方的結果時會發生死鎖,所以MySQL會回滾其中一個。但在我的情況下,MySQL處於自動提交模式,我插入一條觸發觸發器的新記錄。所以我不明白它會導致死鎖的情況。MySQL插入死鎖,引發觸發器
這裏是我的表的模式:
----用戶表----
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
---喜歡錶---
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
我有一個觸發器o ñ喜歡錶,定義如下:
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
隨着做一些簡單的插入:
$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);
}
我得到以下死鎖錯誤日誌:
*** (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)
不應該爲此起來在鎖等待,而不是死鎖?
如何在不重新啓動事務的情況下解決此問題?
你能添加觸發這個的插入嗎?如果你已經將它包裝在一個事務中,請發佈完整的博客 – e4c5
@ e4c5插入代碼僅作爲單個查詢執行。我已經添加了代碼在PHP中插入 – BlackBrain