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mysql事务问题演示

This commit is contained in:
2025-09-01 04:31:37 +08:00
parent 484bd7eb02
commit 71c569c9bd
3 changed files with 76 additions and 0 deletions

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create schema errors;
-- 生成演示用的表
create table errors.customer
(
id int auto_increment
primary key,
username varchar(50) not null
);
create table errors.customer_back
(
id int auto_increment
primary key,
username varchar(50) not null
);

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-- example 1
-- ========== 0 =========
start transaction;
-- A先查询
select *
from customer;
-- ========== 1 =========
-- ========== 2 =========
-- 在 B 添加数据后A 尝试进行修改。奇怪的是,明明查询不到,居然可以修改成功
update customer
set username = 'wuhu'
where id = 1;
-- A 再次查询表中数据,发现修改的那一行现身了
select *
from customer;
commit;
-- ========== 3 =========
-- example 2
-- ========== 0 =========
start transaction;
-- A先查询查到数据如下
-- 1,Fortern
-- 2,Maxin
-- 3,MooGeo
select *
from customer;
-- ========== 1 =========
-- ========== 2 =========
-- A 备份表中数据到另一个表
insert into customer_back select * from customer;
commit;
-- 结果备份表中的数据居然是被其他事物修改的数据!!
select * from customer_back;
-- 为什么会这样?
-- ========== 3 =========
select * from customer where id = 1;

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-- example 1
-- ========== 1 =========
start transaction;
-- B插入数据
insert into customer (id, username)
values (1, 'Fortern'),
(2, 'Maxin'),
(3, 'MooGeo');
commit;
-- ========== 2 =========
-- example 2
-- ========== 1 =========
start transaction;
-- B 修改所有人的name
update customer set username = '哈基米' where true;
commit;
-- ========== 2 =========