spring - 如何使用 Spring 3.2 和 JPA 2.0 动态连接数据库

Question

我正在使用弹簧 3.2.3 和 JPA 2.0。我想基于枚举动态连接我的数据库。我有这样的配置

<bean id="legacyDataSource" name="myLegacyDataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource" lazy-init="true">
    <property name="driverClassName" value="${jdbc.legacy.driverClassName}" />
    <property name="url" value="${jdbc.legacy.url}" />  
    <property name="username" value="${jdbc.legacy.username}" />
    <property name="password" value="${jdbc.legacy.password}" />
</bean>

 <bean id="modernDataSource" name="myModernDataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource" lazy-init="true">
    <property name="driverClassName" value="${jdbc.modern.driverClassName}" />
    <property name="url" value="${jdbc.modern.url}" />  
    <property name="username" value="${jdbc.modern.username}" />
    <property name="password" value="${jdbc.modern.password}" />
</bean>

<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
    <property name="entityManagerFactory" ref="emf"/>
</bean>

<bean id="placeholderProperties" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
    <property name="location" value="classpath:database/jdbc.properties" />
    <property name="systemPropertiesModeName" value="SYSTEM_PROPERTIES_MODE_OVERRIDE" />
    <property name="ignoreUnresolvablePlaceholders" value="true" />
    <property name="order" value="1" />
</bean>

<tx:annotation-driven transaction-manager="transactionManager" />

<bean id="emf" name="myEmf" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" lazy-init="true">
    <property name="dataSource" ref="dataSource" />
    <property name="jpaVendorAdapter" ref="vendorAdaptor" />         
    <property name="packagesToScan" value="com.softech.ls360.integration.regulators.plcb.domain"/>
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</prop>
            <prop key="hibernate.max_fetch_depth">3</prop>
            <prop key="hibernate.jdbc.fetch_size">50</prop>
            <prop key="hibernate.jdbc.batch_size">10</prop>
            <prop key="hibernate.show_sql">true</prop>              
        </props>        
    </property>
</bean>

我将枚举声明为

public enum DatabaseType {

    LEGACY,
    MODERN

} //end of enum DatabaseType

现在我想在我的主类的基础上枚举数据库开关。为此，我认为我需要改变

<property name="dataSource" ref="dataSource" />

在<bean id="emf" name="myEmf" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" lazy-init="true">运行时。我对吗？

有什么办法可以做到这一点，或者spring是否提供了在运行时切换数据库的任何其他方式？

我找到了这篇文章动态数据源路由，但这是一个非常古老的帖子，而且在这篇文章中没有使用 JPA。请帮助

谢谢。

编辑 - - - - - - - - - - - - - - - -

@Lazy
@Service("dbManager") 
@Repository
@Transactional
public class DatabaseManager {

    public enum EnumLmsPlateform {      
        MODERN,
        LEGACY;
    }

    private EnumLmsPlateform lmsPlatform;

    public EnumLmsPlateform getLmsPlatform() {
        return lmsPlatform;
    }

    public void setLmsPlatform(EnumLmsPlateform lmsPlatform) {
        this.lmsPlatform = lmsPlatform;
    }

    @PostConstruct
    public void init() {    
        setLmsPlatform(EnumLmsPlateform.MODERN);    
    }

    @PersistenceContext(unitName="legacy_emf")
    private EntityManager legacyEm;

    @PersistenceContext(unitName="modern_emf")
    private EntityManager ls360Em;

    @SuppressWarnings("unchecked")
    @Transactional(readOnly=true)
    public List<Object> getResultList(String query, Class<?> mappingClass) throws Exception {

        EntityManager em = null;

        if (lmsPlatform == EnumLmsPlateform.LEGACY) {
            em = legacyEm;  
        } else if (lmsPlatform == EnumLmsPlateform.MODERN){ 
            em = ls360Em;   
        }
        Query emQuery = em.createNativeQuery(query, mappingClass);  
        return  emQuery.getResultList();

    } //end of findTraineeFromLegacy()
} //end of class

这是我的弹簧文件

遗留-spring.xml

<bean id="legacyDataSource" name="legacy_DataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource" lazy-init="true">
    <property name="driverClassName" value="${jdbc.legacy.driverClassName}" />
    <property name="url" value="${jdbc.legacy.url}" />  
    <property name="username" value="${jdbc.legacy.username}" />
    <property name="password" value="${jdbc.legacy.password}" />
</bean>

<bean id="legacyTransactionManager" class="org.springframework.orm.jpa.JpaTransactionManager" lazy-init="true">
    <property name="entityManagerFactory" ref="legacyEmf"/>
</bean>

<tx:annotation-driven transaction-manager="legacyTransactionManager"  />

<bean id="legacyEmf" name="legacy_emf" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" lazy-init="true" >
    <property name="dataSource" ref="legacyDataSource" />
    <property name="jpaVendorAdapter" ref="vendorAdaptor" />         
    <property name="packagesToScan" value="com.softech.ls360.integration.regulators.plcb.domain"/>
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</prop>
            <prop key="hibernate.max_fetch_depth">3</prop>
            <prop key="hibernate.jdbc.fetch_size">50</prop>
            <prop key="hibernate.jdbc.batch_size">10</prop>
            <prop key="hibernate.show_sql">true</prop>              
        </props>        
    </property>
</bean>

同一个文件适用于moderns-spring.xml

然后我将这两个文件都包含在我的 maim spring 中

应用程序上下文.xml

<import resource="classpath:database/hibernate-context.xml"/>
<import resource="classpath:database/legacyJpa-context.xml"/>
<import resource="classpath:database/modernJpa-context.xml"/> 

<context:annotation-config/>
....

但是这个设计中有一个流程。两位经理都包括在内。传统与现代。我的意思是当我的 dbManager bean 被初始化时，它需要两者@PersistenceContext(unitName="legacy_emf"), @PersistenceContext(unitName="modern_emf")。应该发生什么，因为我的默认值persistence context是现代的，所以 @PersistenceContext(unitName="modern_emf")在 bean 初始化时应该只有这个持久性上下文可用。然后，如果我设置我的enum to legacy那么 unitName="legacy_emf"这应该是可用的。你得到我了吗？

谢谢

编辑 2：------------------------------------------------------------ ----------------------

我在下面更改了我的 dbManager 类

@Lazy
@Service("dbManager") 
@Repository
@Transactional
public class DatabaseManager {

    public enum EnumLmsPlateform {
        MODERN,
        LEGACY;
    }

    private EnumLmsPlateform lmsPlatform;

    public DatabaseManager() {

    }

    @PostConstruct
    public void init() {
        setLmsPlatform(EnumLmsPlateform.MODERN);
    }


    public EnumLmsPlateform getLmsPlatform() {
        return lmsPlatform;
    }

    public void setLmsPlatform(EnumLmsPlateform lmsPlatform) {
        this.lmsPlatform = lmsPlatform;
    }

    @ModernTx
    public List<Object> getModernResultSet(String query, Class<?> mappingClass) { 
        EntityManager modernEm = getEntityManagerFormBoundTransactionalThread();
        return getResultSet(modernEm, query, mappingClass);

    }

    @LegacyTx 
    public List<Object> getLegacyResultSet(String query, Class<?> mappingClass) { 
        EntityManager legacyEm = getEntityManagerFormBoundTransactionalThread();
        return getResultSet(legacyEm, query, mappingClass);
    } 

    @Transactional 
    public List<Object> getDefaultResultSet(String query, Class<?> mappingClass) { 
        EntityManager defaultEm = getEntityManagerFormBoundTransactionalThread();
        return getResultSet(defaultEm, query, mappingClass);
    }

    private List<Object> getResultSet(EntityManager em, String query, Class<?> mappingClass) {
        Query emQuery = em.createNativeQuery(query, mappingClass);  
        return  emQuery.getResultList();    
    }

    private EntityManager getEntityManagerFormBoundTransactionalThread(){
        Map<Object, Object>  map =   TransactionSynchronizationManager.getResourceMap();   
        for (Object value  :map.values()) {
            if(value instanceof EntityManagerHolder){       
                return ((EntityManagerHolder)value).getEntityManager();

            }  
        } 

        return null;  

    } //end of  getEntityManagerFormBoundTransactionalThread()  

} //end of class

我的测试班

public class TestDynamicJpa {

    String springXmlFile = "classpath:spring/app-context-xml.xml";

    GenericXmlApplicationContext springContext = SpringUtil.loadSpringContext(springXmlFile);

    List<Object> traineeList = getLegacyResultList(springContext);

    private List<Object> getLegacyResultList(GenericXmlApplicationContext springContext) throws Exception {

        DatabaseManager dbManager = springContext.getBean("dbManager", DatabaseManager.class);

        List<Object> resultList = dbManager.getLegacyResultSet(FIND_TRAINEE_LEGACY, LegacyTrainee.class);

        return resultList;

    } //end of getLegacyResultList()

} //end of class TestDynamicJpa

一个问题。当我做

List<Object> resultList = dbManager.getLegacyResultSet(FIND_TRAINEE_LEGACY, LegacyTrainee.class);

然后它LegacyTransactionManager从 Map 中获取，很好。但如果我这样做

List<Object> resultList = dbManager.getDefaultResultSet(FIND_TRAINEE_LEGACY, LegacyTrainee.class);

然后它从 Map 获取哪个事务管理器。Legacy还是Modern？

我也用这个配置检查了它

 <import resource="classpath:database/hibernate-context.xml"/>
 <import resource="classpath:database/legacyJpa-context.xml"/>
 <!--  
 <import resource="classpath:database/ls360Jpa-context.xml"/> 
 -->

即，只导入一个 Jpa 上下文，它工作正常:)。在我之前@PersistenceUnit的配置中，如果我注释掉一个文件，我会收到错误:)。

谢谢：）

score 0 · Accepted Answer

在这种情况下，由于您还有 2 个数据源，您应该再创建 2 个不同名称的 transactionManager。

<bean id="legacyTransactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
    <property name="entityManagerFactory" ref="emfLegacy"/>
</bean>


<bean id="modernTransactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
    <property name="entityManagerFactory" ref="emfModern"/>
</bean>

<bean id="emfLegacy" name="myEmfLegacy" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" lazy-init="true">
    <property name="dataSource" ref="legacyDataSource" />
    <property name="jpaVendorAdapter" ref="vendorAdaptor" />         
    <property name="packagesToScan" value="com.softech.ls360.integration.regulators.plcb.domain"/>
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</prop>
            <prop key="hibernate.max_fetch_depth">3</prop>
            <prop key="hibernate.jdbc.fetch_size">50</prop>
            <prop key="hibernate.jdbc.batch_size">10</prop>
            <prop key="hibernate.show_sql">true</prop>              
        </props>        
    </property>
</bean>   



<bean id="emfModern" name="myEmfModern" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" lazy-init="true">
    <property name="dataSource" ref="modernDataSource" />
    <property name="jpaVendorAdapter" ref="vendorAdaptor" />         
    <property name="packagesToScan" value="com.softech.ls360.integration.regulators.plcb.domain"/>
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</prop>
            <prop key="hibernate.max_fetch_depth">3</prop>
            <prop key="hibernate.jdbc.fetch_size">50</prop>
            <prop key="hibernate.jdbc.batch_size">10</prop>
            <prop key="hibernate.show_sql">true</prop>              
        </props>        
    </property>
</bean>   



 @Target({ElementType.METHOD, ElementType.TYPE})
@Retention(RetentionPolicy.RUNTIME)
@Transactional("modernTransactionManager")
public @interface ModernTx {
}

@Target({ElementType.METHOD, ElementType.TYPE})
@Retention(RetentionPolicy.RUNTIME)
@Transactional("legacyTransactionManager")
public @interface LegacyTx {
}

之后，剩下的就很容易了，只需使用@ModernTx 或@LegacyTx 注释您的bean 的db 方法，如下所示。

public class TransactionalService {

    @ModernTx
    public void setSomething(String name) { 

         EntityManager currentEm = getEntityManagerFormBoundTransactionalThread();
         /*You can do anything with this Modern em*/
    }

    @LegacyTx 
    public void doSomething() { 

         EntityManager currentEm = getEntityManagerFormBoundTransactionalThread();
         /*You can do anything with this Legacy em*/

    } 


    @Transactional 
    public void buySomething() { 

         EntityManager currentEm = getEntityManagerFormBoundTransactionalThread();
         /*You can do anything with this default em*/

    }         




   private getEntityManagerFormBoundTransactionalThread(){

        Map<Object, Object>  map =                    TransactionSynchronizationManager.getResourceMap();
        for (Object value  :map.values()) {
            if(value instanceof EntityManagerHolder){
                            return ((EntityManagerHolder)value).getEntityManager();
                }
            }
            }     
     }

但是，当您希望使用 dafult 数据源时，只需使用 @Transactional，因为您已在上下文中配置了如下默认事务管理器。

spring - 如何使用 Spring 3.2 和 JPA 2.0 动态连接数据库

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