MySQL中是否允许嵌套事务?

InnoDB 支架 SAVEPOINTS .

您可以执行以下操作:

CREATE TABLE t_test (id INT NOT NULL PRIMARY KEY) ENGINE=InnoDB;

START TRANSACTION;

INSERT
INTO    t_test
VALUES  (1);

SELECT  *
FROM    t_test;

 id
---
  1

SAVEPOINT tran2;

INSERT
INTO    t_test
VALUES  (2);

SELECT  *
FROM    t_test;

 id
---
  1
  2

ROLLBACK TO tran2;

SELECT  *
FROM    t_test;

 id
---
  1

ROLLBACK;

SELECT  *
FROM    t_test;

 id
---

从MySQL文档:

不能嵌套事务。这是在发出start transaction语句或其同义词之一时为任何当前事务执行隐式提交的结果。 https://dev.mysql.com/doc/refman/5.7/en/implicit-commit.html

如果您使用PHP,那么您可能会对 https://github.com/Enelar/phpsql 它支持mysql和pgsql,并可扩展到其他连接器。

function TransferMoney()
{ // Nested transaction code could not even know that he nested
  $trans3 = db::Begin();
  if (!db::Query("--Withdraw money from user", [$uid, $amount], true))
    return $trans3->Rollback();
  db::Query("--Deposit money to system");
  return $trans3->Commit();
}

$trans = db::Begin();
db::Query("--Give item to user inventory");
    $trans2 = $trans->Begin();
    $trans2->Query("--Try some actions and then decide to rollback");
    $trans2->Rollback();
// Commit or rollback depending on money transfer result
return $trans->Finish(TransferMoney());