我正在尝试了解JPA中的事务。
假设我已将应用程序部署到具有持久性的glassfish:
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
<persistence-unit name="survex" transaction-type="JTA">
<jta-data-source>jdbc/MyDatabase</jta-data-source>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
</properties>
</persistence-unit>
</persistence>
我正在尝试调试代码嗅探请求到MySQL服务器:
public String create() {
EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory("survex");
EntityManager em = entityManagerFactory.createEntityManager();
Log log = em.find(Log.class, 1);
EntityTransaction transaction = em.getTransaction();
transaction.begin(); // NOTING HAPPENS ...
log.setMessage("xxx");
em.flush(); // OH, HERE IT IS -> SET autocommit=0; UPDATE LOG SET message = 'xxx' WHERE (id = 1)
log.setMessage("aaa");
em.flush(); // OK -> UPDATE LOG SET message = 'aaa' WHERE (id = 1)
transaction.commit(); // NOTHING HAPPENS! WHERE IS COMMIT?!
log = em.find(Log.class, 1);
transaction = em.getTransaction();
transaction.begin(); // STILL NOTHING ...
log.setMessage("555");
em.flush(); // UPDATE LOG SET message = '555' WHERE (id = 1)
log.setMessage("666");
em.flush(); // UPDATE LOG SET message = '666' WHERE (id = 1)
transaction.commit(); // NOTHING!!!
return log.getMessage();
}
// SOMEWHERE OUTSIDE: COMMIT; SET autocommit=1;
为什么?为什么JPA表现得如此奇怪?
我希望在每次transaction.commit()调用时提交,或者我误解了什么?
答案 0 :(得分:1)
我在您的代码中看到一个问题:您将PersistenceContext定义为JTA,但使用RESOURCE_LOCAL事务API(即EntityManager.getTransaction()
)。
当您在Java EE环境中时,只需获取UserTransaction
:@Resource private UserTransaction transaction;
和EntityManager
的注入实例,而不是创建一个实例。您还应该禁用CMT(容器管理的)事务,否则它们会自动打开/关闭。