如果在当前事务的中间执行START TRANSACTION命令,那么当前MySQL事务将如何处理?
如果在当前事务的中间执行STARTTRANSACTION,则将提交并结束当前事务。当前事务中所做的所有数据库更改都将被永久化。STARTTRANSACTION命令将其称为隐式提交。
示例
假设我们在表“marks”中具有以下值
mysql> select * from marks; +------+---------+-----------+-------+ | Id | Name | Subject | Marks | +------+---------+-----------+-------+ | 1 | Aarav | Maths | 50 | | 1 | Harshit | Maths | 55 | | 3 | Gaurav | Comp | 69 | +------+---------+-----------+-------+ 3 rows in set (0.00 sec) mysql> START TRANSACTION; mysql> INSERT INTO Marks Values(4, 'Rahul','History',40); mysql> INSERT INTO Marks Values(5, 'Yashraj','English',48); mysql> START TRANSACTION;
在此示例中,我们可以观察到,在当前事务的中间执行STARTTRANSACTION语句时,它将隐式结束当前事务,并且将提交更改。
mysql> select * from marks; +------+---------+-----------+-------+ | Id | Name | Subject | Marks | +------+---------+-----------+-------+ | 1 | Aarav | Maths | 50 | | 1 | Harshit | Maths | 55 | | 3 | Gaurav | Comp | 69 | | 4 | Rahul | History | 40 | | 5 | Yashraj | English | 48 | +------+---------+-----------+-------+ 5 rows in set (0.00 sec)
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