Question

我正在尝试使用Spring Data JPA，但遇到了保存ManyToOne关系的问题。

@Entity
public class Customer {

 @Id
 @GeneratedValue(strategy=GenerationType.AUTO)
 private Long id;
 private String firstName;
 private String lastName;


 @ManyToOne(cascade=CascadeType.ALL)
 private Address homeAddress;

 protected Customer() {}
 // getters and setters ...
}

许多客户可以拥有相同的地址：

 @Entity
 public class Address {

  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private Long id;

  private String postalCode;
  private String city;
  private String line1;
  private String line2;
  private String country;

  public Address() {
    // TODO Auto-generated constructor stub
  }

  // getters and setter 
}

CustomerRepository

import org.springframework.data.jpa.repository.JpaRepository;

public interface CustomerRepository extends JpaRepository<Customer, Long>       {

List<Customer> findByLastName(String lastName);
List<Customer> findByHomeAddress(Address address);

}

应用

@SpringBootApplication
public class Application {

private static final Logger log = LoggerFactory.getLogger(Application.class);

public static void main(String[] args) {
    SpringApplication.run(Application.class);
}

@Bean
public CommandLineRunner demo(CustomerRepository repository) {
    return (args) -> {



        Address address1 = new Address("07093", "Brooklyn", "129 67th st", null, "USA");
        Address address2 = new Address("03333", "Qeeens", "333 67th st", null, "USA");
        // save a couple of customers
        Customer jack = new Customer("Jack", "Bauer");

        jack.setHomeAddress(address1);
        repository.save(jack);                  
        repository.save(new Customer("Chloe", "O'Brian",address1));

 }
}

所以我尝试使用相同的地址保存两个客户，我得到以下异常：

Caused by: org.springframework.dao.InvalidDataAccessApiUsageException: detached entity passed to persist: hello.Address; nested exception is org.hibernate.PersistentObjectException: detached entity passed to persist: hello.Address
at org.springframework.orm.jpa.vendor.HibernateJpaDialect.convertHibernateAccessException(HibernateJpaDialect.java:299) ~[spring-orm-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.orm.jpa.vendor.HibernateJpaDialect.translateExceptionIfPossible(HibernateJpaDialect.java:244) ~[spring-orm-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.translateExceptionIfPossible(AbstractEntityManagerFactoryBean.java:491) ~[spring-orm-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.dao.support.ChainedPersistenceExceptionTranslator.translateExceptionIfPossible(ChainedPersistenceExceptionTranslator.java:59) ~[spring-tx-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.dao.support.DataAccessUtils.translateIfNecessary(DataAccessUtils.java:213) ~[spring-tx-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.dao.support.PersistenceExceptionTranslationInterceptor.invoke(PersistenceExceptionTranslationInterceptor.java:147) ~[spring-tx-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179) ~[spring-aop-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.data.jpa.repository.support.CrudMethodMetadataPostProcessor$CrudMethodMetadataPopulatingMethodInterceptor.invoke(CrudMethodMetadataPostProcessor.java:133) ~[spring-data-jpa-1.10.5.RELEASE.jar:na]
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179) ~[spring-aop-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.aop.interceptor.ExposeInvocationInterceptor.invoke(ExposeInvocationInterceptor.java:92) ~[spring-aop-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179) ~[spring-aop-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at org.springframework.aop.framework.JdkDynamicAopProxy.invoke(JdkDynamicAopProxy.java:213) ~[spring-aop-4.3.4.RELEASE.jar:4.3.4.RELEASE]
at com.sun.proxy.$Proxy60.save(Unknown Source) ~[na:na]
at hello.Application.lambda$0(Application.java:35) [classes/:na]
at org.springframework.boot.SpringApplication.callRunner(SpringApplication.java:800) [spring-boot-1.4.2.RELEASE.jar:1.4.2.RELEASE]
... 6 common frames omitted

如何解决此问题？您会在this git location

看到完整的代码

Answer 1

添加关系的另一面，使其成为双向的。在Address班，

@OneToMany(mappedBy = "homeAddress", fetch = FetchType.LAZY)
private Set<Customer> customers

编辑：以上建议无法解决问题。你的问题让我很好奇，我在当地检查了一下。当您在每个事务上运行这些操作时，实体管理器会在第一次保存后分离保存的address1（实体管理器/会话在每次存储库交互后关闭）。

例如，如果您将所有操作包装在一个使实体管理器保持打开的事务中（通常在服务上完成，具体取决于您的回滚逻辑），您的方法将起作用我使用以下代码进行了测试，但它确实有效。

    package hello;

    import javax.transaction.Transactional;

    import org.springframework.beans.factory.annotation.Autowired;
    import org.springframework.stereotype.Service;

    @Service
    public class CustomerService {
        @Autowired
        private CustomerRepository customerRepository;

        @Transactional
        public void store(){
            Address address1 = new Address("07093", "Brooklyn", "129 67th st", null, "USA");
            Address address2 = new Address("03333", "Qeeens", "333 67th st", null, "USA");
            // save a couple of customers
            Customer jack = new Customer("Jack", "Bauer");

            jack.setHomeAddress(address1);

            customerRepository.save(jack);                  
            customerRepository.save(new Customer("Chloe", "O'Brian", address1));
        }
    }

另外，我必须从pom.xml中删除 naryana 启动器

Answer 2

首先使用addressRepository保留您的地址，然后参考它。

试试这个

Address address1 = new Address("07093", "Brooklyn", "129 67th st", null, "USA");
Address address2 = new Address("03333", "Qeeens", "333 67th st", null, "USA");
address1 = addressRepository.save(address1); 
address2 = addressRepository.save(address2);

// save a couple of customers
Customer jack = new Customer("Jack", "Bauer");
jack.setHomeAddress(address1);
repository.save(jack);                  
repository.save(new Customer("Chloe", "O'Brian",address1));

PS：我没有对此进行测试，但这应该可行。

Answer 3

所以这是对我实际工作的总结：

1-使用CustomerRepository和AddressRepository作为@Abdullah Wasi建议。这个工作有点长，我有cascade = CascadeType.MERGE作为级联类型：

@ManyToOne(cascade=CascadeType.MERGE)
private Address homeAddress;

2-使用CustomerRepository并根据@isah answer从@Transactional服务执行操作。该选项也有效，因为我将cascade = CascadeType.ALL作为级联类型：

@ManyToOne(cascade=CascadeType.MERGE)
private Address homeAddress;

服务

  package hello;

import javax.transaction.Transactional;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

@Service
public class CustomerService {
    @Autowired
    private CustomerRepository customerRepository;

    @Transactional
    public void store(){
        Address address1 = new Address("07093", "Brooklyn", "129 67th st", null, "USA");
        Address address2 = new Address("03333", "Qeeens", "333 67th st", null, "USA");
        // save a couple of customers
        Customer jack = new Customer("Jack", "Bauer");

        jack.setHomeAddress(address1);

        customerRepository.save(jack);                  
        customerRepository.save(new Customer("Chloe", "O'Brian", address1));
    }
}

我认为@isah是更优化的解决方案，因为我的操作最终将在服务类中完成。

带有ManyToOne保存操作的Spring Data JPA正在抛出org.hibernate.PersistentObjectException

3 个答案: