找不到 MySQL 丢失更新问题的解决方案

Posted 2023-03-23

【中文标题】找不到 MySQL 丢失更新问题的解决方案

【英文标题】：Cannot find solution to MySQL Lost Update Problem

【发布时间】：2021-07-07 02:41:07

【问题描述】：

我正在尝试修复 mysql 中的丢失更新问题，其中两个会话不会相互更新。基本上，问题如下：

A owes B $30. Later, A owes B another $20. In the end A needs to owe both $30 and $20, where A eventually has 50 and B has 150.

我打开了两个终端会话，但不确定如何解决并发问题。我已经搜索了所有内容以找到解决方案，但没有运气，我怎么能让 A 和 B 数字在两个会话中得到相同的结果。

以下是我从两个终端会话中获取的分步过程。

会话 A：

UPDATE accounts SET balance = 100;
SELECT * FROM accounts;

START TRANSACTION;

SELECT balance INTO @user1_balance from accounts where id = 1;
SELECT balance INTO @user2_balance from accounts where id = 2;

SELECT @user1_balance, @user2_balance;

会话 B：

START TRANSACTION;

SELECT balance INTO @user1_balance from accounts where id = 1;
SELECT balance INTO @user2_balance from accounts where id = 2;

SELECT @user1_balance, @user2_balance;

会话 A：

UPDATE accounts 
SET balance = @user1_balance - 30 
WHERE id = 1;

UPDATE accounts 
SET balance = @user2_balance + 30 
WHERE id = 2;

COMMIT;

SELECT * FROM accounts;

会话 B：

SELECT * FROM accounts;

会话 B：

UPDATE accounts 
SET balance = @user1_balance - 20 
WHERE id = 1;

UPDATE accounts 
SET balance = @user2_balance + 20 
WHERE id = 2;

COMMIT;

SELECT * FROM accounts;

如果我运行这些

【参考方案1】：

鉴于您的设置方式，丢失更新是不可避免的。问题是将余额选择为一个变量，然后用于更新表中的余额。

这在生产环境中不会发生，因为不需要为以后存储余额。

如果您将更新作为事务的一部分直接应用，那么一切都应该是一致的：

A 航站楼：

START TRANSACTION;
UPDATE accounts set balance = balance -30 where id = 1;
UPDATE accounts set balance = balance +30 where id = 2;

B 航站楼：

START TRANSACTION;
UPDATE accounts set balance = balance -20 where id = 1;
-- This update is blocked by the outstanding COMMIT on terminal A. Terminal B waits

A 航站楼：

COMMIT;

B 航站楼：

-- The COMMIT on Terminal A unlocks the rows, so we can complete our update.
UPDATE accounts set balance = balance +20 where id = 2;
COMMIT;

现在两个终端显示相同的结果：

select * from accounts;
+----+---------+
| id | balance |
+----+---------+
|  1 |   50.00 |
|  2 |  150.00 |
+----+---------+

【参考方案2】：

您不必运行SELECT 来分配用户余额。可以在UPDATE完成

UPDATE accounts 
SET balance = balance  - 20 
WHERE id = 1;

注意balance = balance - 20。