我试图设置基本的JPA插入测试。但是在数据库中没有保存任何内容。 DB是Postgresql。 Hibernate被用作持久性提供者。使用Spring JPA持久/承诺不能在测试环境中工作JUnit
非常感谢提前。
@Entity
public class Currency {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
protected Integer id;
@Column
private String code;
@Column
private String name;
...
}
的CRUD类:所有测试
@Repository
@Scope(BeanDefinition.SCOPE_PROTOTYPE)
@Transactional(propagation = Propagation.REQUIRED)
public class CRUDServiceBean implements CRUDService {
@PersistenceContext(type = PersistenceContextType.EXTENDED)
private EntityManager entityManager;
public EntityManager getEntityManager() {
return entityManager;
}
public <T extends BaseEntity> T persistAndCommit(T t) {
entityManager.persist(t);
entityManager.refresh(t);
entityManager.getTransaction().commit();
return t;
}
...
...
}
基类:
@RunWith(SpringJUnit4ClassRunner.class)
@Configurable(autowire = Autowire.BY_NAME)
@ContextConfiguration(locations = { "classpath:context-test.xml" })
public class BaseTest {
}
测试类:
public class CurrencyCreateTest extends BaseTest {
@Autowired
CRUDService crudService;
@Test
@Transactional(propagation = Propagation.REQUIRES_NEW)
public void createCurrency() throws Exception {
Currency currency = new Currency();
currency.setCode("EUR");
currency.setName("Euro");
currency = crudService.persistAndCommit(currency);
}
}
上下文的test.xml:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:task="http://www.springframework.org/schema/task"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-4.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-4.0.xsd">
<context:component-scan base-package="com.chartinvest"/>
<bean id="contextApplicationContextProvider" class="com.chartinvest.util.ApplicationContextProvider"></bean>
<!-- the parent application context definition for the springapp application -->
<!-- dataSource -->
<bean id="dataSourceFinance" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName"><value>org.postgresql.Driver</value></property>
<property name="url"><value>jdbc:postgresql://localhost/db_finance_test</value></property>
<property name="username"><value>postgres</value></property>
<property name="password"><value>xxxxxxxx</value></property>
</bean>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="persistenceUnitName" value="mypersistenceunit" />
<property name="dataSource" ref="dataSourceFinance" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="true" />
<property name="databasePlatform" value="org.hibernate.dialect.PostgreSQLDialect" />
</bean>
</property>
<property name="jpaPropertyMap">
<map>
<entry key="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect"/>
</map>
</property>
</bean>
<!-- Enable the configuration of transactional behavior based on annotations -->
<tx:annotation-driven transaction-manager="transactionManager"/>
</beans>
尝试添加'entityManager.merge(T)'之前,你的'entityManager.persist(T)'在'persistAndCommit' – dazito
是什么意思它不工作?你有错误吗?或者你期望有一些结果,而且它是不同的?请分享预期与实际行为。 –