java – Spring @Transactional提交失败; Deby Eclipselink

以下是 spring配置

日期来源

<bean id="dataSource" class="org.apache.commons.dbcp2.BasicDataSource">
    <property name="driverClassName" value="${rwos.dataSource.driverClassName}" />
    <property name="url" value="${rwos.dataSource.url}" />
    <property name="username" value="${rwos.dataSource.user}" />
    <property name="password" value="${rwos.dataSource.password}" />
 </bean>

实体经理配置

<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context="http://www.springframework.org/schema/context"
       xmlns:tx="http://www.springframework.org/schema/tx"
       xsi:schemaLocation="
            http://www.springframework.org/schema/context
            http://www.springframework.org/schema/context/spring-context.xsd
            http://www.springframework.org/schema/beans
            http://www.springframework.org/schema/beans/spring-beans.xsd
            http://www.springframework.org/schema/tx
            http://www.springframework.org/schema/tx/spring-tx.xsd">

  <bean id="persistenceUnitManager" class="org.springframework.data.jpa.support.MergingPersistenceUnitManager">
        <property name="defaultDataSource" ref="dataSource"/>
  </bean>

    <bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
        <property name="persistenceUnitManager" ref="persistenceUnitManager"/>
        <property name="persistenceUnitName" value="com.retailwave.rwos_rwos-data-pojo_jar_4.0.0PU"/>
  </bean>

  <bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager"  lazy-init="true">
                <property name="entityManagerFactory" ref="entityManagerFactory"/>
                <property name="dataSource" ref="dataSource" />
  </bean>
    <tx:annotation-driven transaction-manager="transactionManager" proxy-target-class="true"/>
</beans>

以下是用于持久化实体的代码段

@Singleton
@Component
public class RWTransactionDao {

@PersistenceContext(type = PersistenceContextType.EXTENDED)
private EntityManager em;

 @Override
protected EntityManager getEntityManager() {
    return em;
}

@Transactional
public void createOrderTxns(RWRetailTransaction peTxn,RWRetailTransaction fcTxn) {
    create(peTxn);
    create(fcTxn);
    log.debug("Committed Transaction : {} ",peTxn.displayString());
    log.debug("Committed Transaction : {} ",fcTxn.displayString());
 }

 @Transactional
 public void create(T entity) {
   getEntityManager().persist(entity);
 }

}

分类：

@Component
@Path("/")
public class RWTransactionREST {

   @Inject
   private RWTransactionDao rWTransactionDao;

  @POST
  @Produces(MediaType.APPLICATION_JSON)
  @Path("txns/sales")
  public RWResponse createPurchaseTransaction(RWRetailTransaction peTxn,RWRetailTransaction fcTxn) {
    rWTransactionDao.createOrderTxns(peTxn,fcTxn);
    builder.status(RWStatus.OK);
    RWResponse response = builder.build();
    log.info("Returning.. {}",response);
    return response; 
   }

  }

记录消息：

2017-06-14 10:49:51,453 DEBUG [qtp592179046-13] - Committed Transaction : RWRetailTransaction[609,CU_ORD,RTCO-609-17-11193,2017-06-14 10:49:51.431] 
2017-06-14 10:49:51,453 DEBUG [qtp592179046-13]  - Committed Transaction : RWRetailTransaction[509,RTCO-509-17-11193,2017-06-14 10:49:51.444]
2017-06-14 10:49:51,463 INFO  [qtp592179046-13] RWTransactionREST - Returning.. Response[1000:Order has been created successfully,Transaction Id : RTCO-609-17-11193]

一段时间后,在其他一些REST实现方法中,同一个RWTransactionDao中发生了以下错误

2017-06-14 10:51:24,199 ERROR [qtp592179046-16] com.retailwave.rwos.compartment.rest.exception.RWCompartmentRestExceptionMapper - Exception caught at Mapper : Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.6.4.v20160829-44060b6): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLTransactionRollbackException: A lock could not be obtained within the time requested

由于此错误,先前的提交已回滚但不应该回滚.

不确定导致此回滚的原因.

解决方法

所以对于你的问题,我的理解是你试图在一个事务中将两条记录INSERT到Derby,所以当你执行create(peTxn); derby锁定单行并保持每一行直到事务提交[lockA],然后你尝试执行create(fcTxn),它会尝试获取另一个单行锁来INSERT一条新记录,但实际上该事务不是因此锁仍然锁定[lockA].

所以解决方案是

调用getEntityManager().刷新事务以在第一步完成时提交事务.这将把SQL INSERT的重点放在数据库上.

并建议在服务层使用@Transactional.

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