一、问题起源
在MySQL的官方文档中有明确的说明不支持嵌套事务:
代码如下:
Transactions cannot be nested. This is a consequence of the implicit commit performed for any current transaction when you issue a START TRANSACTION statement or one of its synonyms.
但是在我们开发一个复杂的系统时难免会无意中在事务中嵌套了事务,比如A函数调用了B函数,A函数使用了事务,并且是在事务中调用了B函数,B函数也有一个事务,这样就出现了事务嵌套。这时候其实A的事务就意义不大了,为什么呢?上面的文档中就有提到,简单的翻译过来就是:
代码如下:
当执行一个START TRANSACTION指令时,会隐式的执行一个commit操作。
所以我们就要在系统架构层面来支持事务的嵌套。所幸的是在一些成熟的ORM框架中都做了对嵌套的支持,比如doctrine或者laravel。接下来我们就一起来看下这两个框架是怎样来实现的。
友情提示,这两个框架的函数和变量的命名都比较的直观,虽然看起来很长,但是都是通过命名就能直接得知这个函数或者变量的意思,所以不要一看到那么一大坨就被吓到了 :)
二、doctrine的解决方案
首先来看下在doctrine中创建事务的代码(干掉了不相关的代码):
代码如下:
public function beginTransaction()
{
++$this->_transactionNestingLevel;
if ($this->_transactionNestingLevel == 1) {
$this->_conn->beginTransaction();
} else if ($this->_nestTransactionsWithSavepoints) {
$this->createSavepoint($this->_getNestedTransactionSavePointName());
}
}
这个函数的第一行用一个_transactionNestingLevel来标识当前嵌套的级别,如果是1,也就是还没有嵌套,那就用默认的方法执行一下START TRANSACTION就ok了,如果大于1,也就是有嵌套的时候,她会帮我们创建一个savepoint,这个savepoint可以理解为一个事务记录点,当需要回滚时可以只回滚到这个点。
然后看下rollBack函数:
代码如下:
public function rollBack()
{
if ($this->_transactionNestingLevel == 0) {
throw ConnectionException::noActiveTransaction();
}
if ($this->_transactionNestingLevel == 1) {
$this->_transactionNestingLevel = 0;
$this->_conn->rollback();
$this->_isRollbackOnly = false;
} else if ($this->_nestTransactionsWithSavepoints) {
$this->rollbackSavepoint($this->_getNestedTransactionSavePointName());
--$this->_transactionNestingLevel;
} else {
$this->_isRollbackOnly = true;
--$this->_transactionNestingLevel;
}
}
可以看到处理的方式也很简单,如果level是1,直接rollback,否则就回滚到前面的savepoint。
然后我们继续看下commit函数:
代码如下:
public function commit()
{
if ($this->_transactionNestingLevel == 0) {
throw ConnectionException::noActiveTransaction();
}
if ($this->_isRollbackOnly) {
throw ConnectionException::commitFailedRollbackOnly();
}
if ($this->_transactionNestingLevel == 1) {
$this->_conn->commit();
} else if ($this->_nestTransactionsWithSavepoints) {
$this->releaseSavepoint($this->_getNestedTransactionSavePointName());
}
--$this->_transactionNestingLevel;
}
算了,不费口舌解释这段了吧 :)
三、laravel的解决方案
laravel的处理方式相对简单粗暴一些,我们先来看下创建事务的操作:
代码如下:
public function beginTransaction()
{
++$this->transactions;
if ($this->transactions == 1)
{
$this->pdo->beginTransaction();
}
}
感觉如何?so easy吧?先判断当前有几个事务,如果是第一个,ok,事务开始,否则就啥都不做,那么为啥是啥都不做呢?继续往下看rollBack的操作:
代码如下:
public function rollBack()
{
if ($this->transactions == 1)
{
$this->transactions = 0;
$this->pdo->rollBack();
}
else
{
--$this->transactions;
}
}
明白了吧?只有当当前事务只有一个的时候才会真正的rollback,否则只是将计数做减一操作。这也就是为啥刚才说laravel的处理比较简单粗暴一些,在嵌套的内层里面实际上是木有真正的事务的,只有最外层一个整体的事务,虽然简单粗暴,但是也解决了在内层新建一个事务时会造成commit的问题。原理就是这个样子了,为了保持完整起见,把commit的代码也copy过来吧!
代码如下:
public function commit()
{
if ($this->transactions == 1) $this->pdo->commit();
--$this->transactions;
}

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MySQL is suitable for beginners because: 1) easy to install and configure, 2) rich learning resources, 3) intuitive SQL syntax, 4) powerful tool support. Nevertheless, beginners need to overcome challenges such as database design, query optimization, security management, and data backup.

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