mysql - 为什么 MySQL 显式谓词锁定不允许在谓词锁之外插入语句

Question

假设我们有以下数据库表:

create table department (
    id bigint not null, 
    budget bigint not null, 
    name varchar(255), 
    primary key (id)
) ENGINE=InnoDB

create table employee (
    id bigint not null, 
    name varchar(255), 
    salary bigint not null, 
    department_id bigint, primary key (id)
) ENGINE=InnoDB

alter table employee 
add constraint FK_department_id 
foreign key (department_id) 
references department (id)

我们有 2 个部门:

insert into department (name, budget, id) 
values ('Hypersistence', 100000, 1)

insert into department (name, budget, id) 
values ('Bitsystem', 10000, 2)

第一个部门有 3 个员工:

insert into employee (department_id, name, salary, id) 
values (1, 'John Doe 0', 30000, 0)

insert into employee (department_id, name, salary, id) 
values (1, 'John Doe 1', 30000, 1)

insert into employee (department_id, name, salary, id) 
values (1, 'John Doe 2', 30000, 2)

假设我们有两个并发用户:Alice 和 Bob。

首先，Alice 锁定属于第一个 department 的所有员工，并获得该特定 department 的工资总和:

SELECT * 
FROM employee 
WHERE department_id = 1 
FOR UPDATE

SELECT SUM(salary) 
FROM employee 
where department_id = 1 

现在，与此同时，预计 Bob 不能使用相同的 department_id 插入新的 employee:

insert into employee (department_id, name, salary, id) 
values (1, `Carol`, 9000, 4)

com.mysql.jdbc.exceptions.jdbc4.MySQLTransactionRollbackException: 
    Lock wait timeout exceeded; try restarting transaction

因此，锁阻止了 Bob 对同一个谓词发出插入。

但是，即使 Bob 尝试在不同的 department 中插入一个 employee，也会抛出相同的异常:

insert into employee (department_id, name, salary, id) 
values (2, `Dave`, 9000, 5)

com.mysql.jdbc.exceptions.jdbc4.MySQLTransactionRollbackException: 
    Lock wait timeout exceeded; try restarting transaction

最后一个插入语句使用第二个 department_id，因此该行不应与我们为其获取谓词锁的 select 语句重叠。

为什么MySQL会阻止与第一个事务获取的谓词锁不重叠的第二个insert？

在 SQL Server 上也可以观察到相同的行为。

更新

当将隔离级别更改为 READ_COMMITTED 时，谓词锁不会阻止 Bob 发出的两个插入语句中的任何一个。

如果考虑到 this Percona blog post 中的以下陈述，就可以解释这一点:

In REPEATABLE READ every lock acquired during a transaction is held for the duration of the transaction.

In READ COMMITTED the locks that did not match the scan are released after the STATEMENT completes.

但是，找出谓词锁定为何像在可重复读上那样工作仍然很有趣。

Answer 1

SELECT FOR UPDATE 锁定在 1 和 employee 表中的下一个值之间。由于没有下一个值，它一直锁定到 supremum pseudo-record。这可以在 information_schema.innodb_locks 中看到:

mysql> select * from innodb_locks;
+----------------+-------------+-----------+-----------+-------------------+------------+------------+-----------+----------+------------------------+
| lock_id        | lock_trx_id | lock_mode | lock_type | lock_table        | lock_index | lock_space | lock_page | lock_rec | lock_data              |
+----------------+-------------+-----------+-----------+-------------------+------------+------------+-----------+----------+------------------------+
| 28275:1448:3:1 | 28275       | X         | RECORD    | `test`.`employee` | PRIMARY    |       1448 |         3 |        1 | supremum pseudo-record |
| 28273:1448:3:1 | 28273       | X         | RECORD    | `test`.`employee` | PRIMARY    |       1448 |         3 |        1 | supremum pseudo-record |
+----------------+-------------+-----------+-----------+-------------------+------------+------------+-----------+----------+------------------------+
2 rows in set, 1 warning (0.00 sec)

如果您稍微更改测试用例，以便在 dept-id=2 的员工中有一行，然后尝试为 dept-id=3 添加一个员工，它将起作用。示例:

create table department (
    id bigint not null, 
    budget bigint not null, 
    name varchar(255), 
    primary key (id)
) ENGINE=InnoDB;

create table employee (
    id bigint not null, 
    name varchar(255), 
    salary bigint not null, 
    department_id bigint, primary key (id)
) ENGINE=InnoDB;


alter table employee 
add constraint FK_department_id 
foreign key (department_id) 
references department (id);


insert into department (name, budget, id) 
values ('Hypersistence', 100000, 1);

insert into department (name, budget, id) 
values ('Bitsystem', 10000, 2);

insert into department (name, budget, id) 
values ('XX', 10000, 3);


insert into employee (department_id, name, salary, id) 
values (1, 'John Doe 0', 30000, 0);

insert into employee (department_id, name, salary, id) 
values (1, 'John Doe 1', 30000, 1);

insert into employee (department_id, name, salary, id) 
values (2, 'John Doe 2', 30000, 2);


start transaction;

SELECT * 
FROM employee 
WHERE department_id = 1 
FOR UPDATE;


# new session

insert into employee (department_id, name, salary, id) 
values (3, 'Dave', 9000, 5)