jan*_*ola 5 junit spring transactions web-applications spring-data-jpa
我的交易有点问题.我使用Spring 3.1.1.RELEASE,Spring Data 1.0.3.RELEASE JPA和Hibernate提供程序.当我开始一个junit测试时,一个用@Transactional它注释的方法似乎很好,但是当我启动一个完整的应用程序时,没有错误但是事务不起作用.这是我的配置和示例代码:
applicationContext.xml中
<context:annotation-config />
<context:component-scan base-package="com.sheedo.upload" />
<jpa:repositories base-package="com.sheedo.upload.repository" />
<tx:annotation-driven transaction-manager="transactionManager" />
<bean
class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="locations">
<list>
<value>classpath*:messages/*.properties</value>
<value>classpath*:*.properties</value>
</list>
</property>
</bean>
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="persistenceUnitName" value="persistenceUnit" />
<property name="dataSource" ref="dataSource" />
</bean>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<bean id="dataSource" class="com.mysql.jdbc.jdbc2.optional.MysqlDataSource">
<property name="url" value="${jdbc.url}" />
<property name="user" value="${jdbc.username}" />
<property name="password" value="${jdbc.password}" />
</bean>
Run Code Online (Sandbox Code Playgroud)
persistence.xml中
<persistence-unit name="persistenceUnit" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLInnoDBDialect" />
<property name="hibernate.hbm2ddl.auto" value="update" />
<property name="hibernate.ejb.naming_strategy" value="org.hibernate.cfg.ImprovedNamingStrategy" />
<property name="hibernate.connection.charSet" value="UTF-8" />
<property name="hibernate.show_sql" value="true" />
</properties>
</persistence-unit>
Run Code Online (Sandbox Code Playgroud)
UserRepository.java
public interface UserRepository扩展CrudRepository <User,Long> {}
UserServiceImpl.java
@Service("userService")
public class UserServiceImpl implements UserService {
@Autowired
private UserRepository userRepository;
@Override
@Transactional
public void addUser(String name, String surname) {
User u = new User(name, surname);
userRepository.save(u);
throw new RuntimeException(); // to invoke a rollback
}
}
Run Code Online (Sandbox Code Playgroud)
UserServiceTest.java
@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = { "classpath:/META-INF/spring/root-context.xml" })
public class UserServiceTest {
Logger log = LoggerFactory.getLogger(getClass());
@Autowired
private UserService userService;
@Test
public void testUserAdd() {
userService.addUser("John", "Doe");
}
}
Run Code Online (Sandbox Code Playgroud)
在这种JUnit测试的情况下,虽然服务方法注释了,但事务不起作用@Transactional.当我将这个注释添加到testUserAdd()方法时,我在控制台中得到它:
2012-05-17 11:17:54,208 INFO [org.springframework.test.context.transaction.TransactionalTestExecutionListener] - Rolled back transaction after test execution for test context [[TestContext@23ae2a testClass = UserRepositoryTest, testInstance = com.sheedo.upload.repository.UserRepositoryTest@7f52c1, testMethod = testUserAdd@UserRepositoryTest, testException = java.lang.RuntimeException, mergedContextConfiguration = [MergedContextConfiguration@111fd28 testClass = UserRepositoryTest, locations = '{classpath:/META-INF/spring/root-context.xml}', classes = '{}', activeProfiles = '{}', contextLoader = 'org.springframework.test.context.support.DelegatingSmartContextLoader']]]
Run Code Online (Sandbox Code Playgroud)
这是正确的,我想.那么,@Transactional注释怎么可能只在Junit测试类中起作用,而在其他的spring bean中却不行?
我的理论是以SpringJUnit4ClassRunner某种方式提供这种交易.我的弹簧配置中有什么问题导致交易在我的应用程序中不起作用但仅在Junit测试类中有效吗?appContext中缺少什么?
编辑: 日志:
2012-05-17 12:46:10,770 DEBUG [org.springframework.orm.jpa.JpaTransactionManager] - Creating new transaction with name [org.springframework.data.jpa.repository.support.SimpleJpaRepository.save]: PROPAGATION_REQUIRED,ISOLATION_DEFAULT; ''
2012-05-17 12:46:10,770 DEBUG [org.springframework.orm.jpa.JpaTransactionManager] - Opened new EntityManager [org.hibernate.ejb.EntityManagerImpl@e4080] for JPA transaction
2012-05-17 12:46:10,979 DEBUG [org.springframework.orm.jpa.JpaTransactionManager] - Not exposing JPA transaction [org.hibernate.ejb.EntityManagerImpl@e4080] as JDBC transaction because JpaDialect [org.springframework.orm.jpa.DefaultJpaDialect@1f87491] does not support JDBC Connection retrieval
Hibernate: insert into user (name, surname) values (?, ?)
2012-05-17 12:46:11,062 DEBUG [org.springframework.orm.jpa.JpaTransactionManager] - Initiating transaction commit
2012-05-17 12:46:11,062 DEBUG [org.springframework.orm.jpa.JpaTransactionManager] - Committing JPA transaction on EntityManager [org.hibernate.ejb.EntityManagerImpl@e4080]
2012-05-17 12:46:11,142 DEBUG [org.springframework.orm.jpa.JpaTransactionManager] - Closing JPA EntityManager [org.hibernate.ejb.EntityManagerImpl@e4080] after transaction
2012-05-17 12:46:11,142 DEBUG [org.springframework.orm.jpa.EntityManagerFactoryUtils] - Closing JPA EntityManager
Run Code Online (Sandbox Code Playgroud)
我有完全相同的问题。<tx:annotation-driven/>另外,阅读在我的网络配置中添加标签的解决方案(spring-servlet.xml而不是applicationContext.xml),并且为我工作。
但我认为这不是一个好的解决方案,所以我试图理解为什么会发生这种情况......
好吧,事实证明,<context:component-scan>我的标签中spring-servlet.xml也包含了@Service其扫描中的类(base-package规范太笼统)。这很奇怪,因为我有一个include-filter引用注释的地方@Controller......但无论如何,似乎 Web 层的应用程序上下文是创建实例的@Service应用程序上下文,而不是为其创建的应用程序上下文applicationContext.xml- 这是定义业务层——并且由于前者没有启用事务性......我没有任何事务。
解决方案(好的一个):更好(更具体)的component-scan配置spring-servlet.xml
| 归档时间: |
|
| 查看次数: |
5025 次 |
| 最近记录: |