java LazyInitializationException：无法初始化代理 - 没有会话
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LazyInitializationException: could not initialize proxy - no Session
提问by Shuai Junlan
I use spring-data-jpawith spring-boot(v2.0.0.RELEASE), I just wrote a CRUD demo on MySQL, but an exception occurs during runtime, source code as follows:
我用spring-data-jpawith spring-boot(v2.0.0.RELEASE)，刚在MySQL上写了一个CRUD的demo，但是运行时出现异常，源码如下：
Source code
源代码
User.java
用户.java
@Entity
public class User implements Serializable {
private static final long serialVersionUID = 1L;
@Id
private Integer id;
private String username;
private String password;
...getter&setter
}
UserRepository.java
用户存储库.java
public interface UserRepository extends JpaRepository<User, Integer> {
}
UserServiceTest.java
用户服务测试.java
@RunWith(SpringRunner.class)
@SpringBootTest
public class UserServiceTest {
@Autowired
private UserRepository userRepository;
@Test
public void getUserById() throws Exception{
userRepository.getOne(1);
}
}
application.yml
应用程序.yml
spring:
datasource:
username: ***
password: ***
driver-class-name: com.mysql.jdbc.Driver
url: ********
thymeleaf:
cache: false
jpa:
show-sql: true
hibernate:
ddl-auto: update
Exception details
异常详情
org.hibernate.LazyInitializationException: could not initialize proxy - no Session
at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:155)
at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:268)
at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:73)
at cn.shuaijunlan.sdnsecuritysystem.domain.po.User_$$_jvstc90_0.getUsername(User_$$_jvstc90_0.java)
at cn.shuaijunlan.sdnsecuritysystem.service.UserServiceTest.getUserById(UserServiceTest.java:33)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at org.junit.runners.model.FrameworkMethod.runReflectiveCall(FrameworkMethod.java:50)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
I try to another method
userRepository.findOne(1), it can run successfully.
我尝试了另一种方法
userRepository.findOne(1)，它可以成功运行。
回答by Cepr0
You can add @Transactionalannotation to your test method to avoid this exception.
您可以@Transactional向测试方法添加注释以避免此异常。
Method getOnereturn the 'reference' (proxy) of the entity which properties can be lazy loaded. See it code- it uses getReferencemethod of EntityManager. From it javadoc:
方法getOne返回可以延迟加载属性的实体的“引用”（代理）。看到它的代码-它使用getReference的方法EntityManager。从它javadoc：
Get an instance, whose state may be lazily fetched.
获取一个实例，其状态可能会被延迟获取。
In Spring the implementation of EntityManageris org.hibernate.internal.SessionImpl- so without the Session the Spring can not get this method.
在 Spring 中的实现EntityManager是org.hibernate.internal.SessionImpl- 所以没有 Session Spring 无法获得这个方法。
To have a session you can just create a transaction...
要进行会话，您只需创建一个事务...
回答by mukesh kumar
Your test should be like this:
你的测试应该是这样的：
@RunWith(SpringRunner.class)
@SpringBootTest
@Transactional
public class QuestionTesting {
@Test
public void test() {
}
}
回答by KayV
In the service class please add a setter for entity manager by using the annoattion @PersistenceContext. To be specific, use
在服务类中，请使用注解@PersistenceContext 为实体管理器添加一个setter。具体来说，使用
@PersistenceContext(type = PersistenceContextType.EXTENDED)
In this way, you can access lazy property.
通过这种方式，您可以访问lazy 属性。
