Question

我正在使用Spring mvc进行项目，我也想使用jpa功能。我有一个3个实体类，它们对应的存储库接口。我在通用服务类中有自己的自动对象。但是，在为控制器中使用的此服务类对象创建bean时，我遇到了问题。 3个模型类是User，Appointment和Notification。存储库接口扩展了CRUDRepository接口服务类：

@Service

public class EHealthService

{

    @Autowired
    UserRepository uRepo;

    @Autowired
    AppointmentRepository aRepo;

    @Autowired
    NotificationRepository nRepo;

    public void registerUser(User u)
    {
        uRepo.save(u);
    }


    public boolean login(User u)
    {
        if(uRepo.findByemail(u.getEmail())!=null)
            return true;
        else
            return false;
    }

    public List<User> getDoctorList()
    {
        return uRepo.findByisdoctor(true);
    }
   // some more functions

}

控制器类：

@Controller

public class EHealthController 

{

    @Autowired
    EHealthService eservice;
    //Some code
}

ehealth-dispacter-servlet.xml文件：

<beans 
xmlns="http://www.springframework.org/schema/beans"
xmlns:mvc="http://www.springframework.org/schema/mvc"
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"
xmlns:jpa="http://www.springframework.org/schema/data/jpa"
xsi:schemaLocation=
"http://www.springframework.org/schema/beans
    http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
    http://www.springframework.org/schema/mvc 
    http://www.springframework.org/schema/mvc/spring-mvc-4.0.xsd
    http://www.springframework.org/schema/context 
    http://www.springframework.org/schema/context/spring-context-4.0.xsd
    http://www.springframework.org/schema/data/jpa
http://www.springframework.org/schema/data/jpa/spring-jpa.xsd
 http://www.springframework.org/schema/tx 
http://www.springframework.org/schema/tx/spring-tx-4.0.xsd">


<context:component-scan base-package="com.cl3.controller" />
<context:component-scan base-package="com.cl3.model" />
<context:component-scan base-package="com.cl3.service" />

<bean     class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <property name="prefix">
        <value>/WEB-INF/pages/</value>
    </property>
    <property name="suffix">
        <value>.jsp</value>
    </property>
</bean>

<bean id="jpaVendorAdapter" class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
    <property name="showSql" value="true"/>
    <property name="generateDdl" value="true"/>
</bean>

<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="dataSource" ref="dataSource"/>
    <property name="jpaVendorAdapter" ref="jpaVendorAdapter"/>

    <property name="packagesToScan" value="com.cl3.model"/>
</bean>

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

 <jpa:repositories base-package="com.cl3.model"/>

<bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource"
    destroy-method="close">
    <property name="driverClassName" value="org.postgresql.Driver" />
    <property name="url" value="jdbc:postgresql://localhost:5432/cl3" />
    <property name="username" value="ucan" />
    <property name="password" value="ucan" />
</bean>

<bean id="eservice" class="com.cl3.service.EHealthService">
    <property name="uRepo" ref="uRepo"></property>
    <property name="nRepo" ref="nRepo"></property>
    <property name="aRepo" ref="aRepo"></property>
</bean>

  <bean id="uRepo"  class="com.cl3.model.UserRepository">
</bean>
 <bean id="nRepo" class="com.cl3.model.NotificationRepository">
</bean>
<bean id="aRepo"  class="com.cl3.model.AppointmentRepository">
</bean>

它说这个类是一个接口。 eservice对象的bean是什么，在dispacter servel xml文件中启用jpa？

谢谢。

Answer 1

如果您使用的是基于spring xml的配置，请将以下bean添加到配置文件中：

 <bean id="entityManagerFactoryBean" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
  <property name="dataSource" ref="dataSource" />
  <!-- This makes /META-INF/persistence.xml is no longer necessary -->
  <property name="packagesToScan" value="com.howtodoinjava.demo.model" />
  <!-- JpaVendorAdapter implementation for Hibernate EntityManager.
       Exposes Hibernate's persistence provider and EntityManager extension interface -->
  <property name="jpaVendorAdapter">
     <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter" />
  </property>
  <property name="jpaProperties">
     <props>
        <prop key="hibernate.hbm2ddl.auto">validate</prop>
        <prop key="hibernate.dialect">org.hibernate.dialect.MySQL5Dialect</prop>
     </props>
  </property>

   <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
  <property name="driverClassName" value="com.mysql.jdbc.Driver" />
  <property name="url" value="jdbc:mysql://localhost:3306/test" />
  <property name="username" value="root" />
  <property name="password" value="password" />

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

通过在xml文件中添加此绑定注释来注册组件扫描：

<context:component-scan base-package="com.mycompany.projectname.demo" />

如果您的项目是spring mvc，请将以下绑定注释添加到xl文件

 <!-- This allow for dispatching requests to Controllers -->
<mvc:annotation-driven />

对于声明式事务管理，在xml文件中添加以下部分。

   <tx:annotation-driven />

如果在spring中启用了组件扫描功能，基本上不需要在xml配置文件中添加服务bean。添加所需的依赖项以将jpa与spring集成。请参阅此链接将为您提供更多帮助：

https://howtodoinjava.com/jpa/spring-hibernate-jpa-configuration-example/

Dispacter Servelt xml文件

1 个答案: