Java Spring 具有@OneToMany 关系的数据@Query returns 无结果

Question

我有以下实体：

@Entity
public class Customer extends BaseEntity {

    private String firstname;
    private String lastname;
    @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL)
    private Set<Address> addresses;

    ...

@Entity
public class Address extends BaseEntity {

    private String street;
    private String houseNumber;
    private String zipCode;
    private String city;
    @ManyToOne
    private Customer customer;

    ...

以及以下存储库界面class：

@Repository
public interface CustomerRepository extends CrudRepository<Customer, Long> {

    @Query("select c from Customer c join c.addresses a where (a.city = :cityName)")
    List<Customer> findByCity(@Param("cityName")String city);

}

现在，我正在尝试运行以下集成测试，但它失败了，我完全不知道为什么。不幸的是，我是 Spring 的初学者，我正在努力学习它 ;-)

@Test
public void testFindCustomerByCity() {
    Customer customer = new Customer("Max", "Tester");
    Address address = new Address("Street", "1", "12345", "City");
    HashSet<Address> addresses = new HashSet<Address>();
    addresses.add(address);
    customer.setAddresses(addresses);
    Customer savedCustomer = customerRepository.save(customer);
    Assert.assertTrue(savedCustomer.getId() > 0);

    List<Customer> customerList = customerRepository.findByCity("City");
    Assert.assertThat(customerList.size(), is(1));
}

错误信息是：

java.lang.AssertionError: Expected: is <1> but: was <0>

为什么结果是空的。我的测试设置错了吗？实体关系？没关系，如果你能帮助我。

Answer 1

您在 Customer 实体的 addresses 字段上有 @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL)。这基本上意味着该关系由 Address 实体中 customer 字段中的值管理。

在您的测试代码中，您只设置了客户的地址，而不是地址上的客户。它仍然是空的，所以数据库中可能有 2 条记录，但没有关系。因此，您的查询不会返回任何内容。

像使用 setAddresses 那样设置集合是在 JPA 环境中做事的一种非常糟糕的方式（当您在已经存在的实例上执行此操作时，您将覆盖持久性集合）。删除 setAddresses 方法并在 Customer 上创建一个 addAddress 方法。

@Entity
public class Customer extends BaseEntity {

    private String firstname;
    private String lastname;

    @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL)
    private final Set<Address> addresses = new HashSet<Address>();

    // No setter, only a getter which returns an immutable collection
    public Set<Address> getAddresses() {
        return Collections.unmodifiableSet(this.addresses);
    }

    public void addAddress(Address address) {
        address.setCustomer(this);
        this.addresses.add(address);
    }

}

这也会稍微清理一下您的测试代码。

@Test
public void testFindCustomerByCity() {
    Customer customer = new Customer("Max", "Tester");
    customer.addAddress(new Address("Street", "1", "12345", "City"));
    Customer savedCustomer = customerRepository.save(customer);

    Assert.assertTrue(savedCustomer.getId() > 0);

    List<Customer> customerList = customerRepository.findByCity("City");
    Assert.assertThat(customerList.size(), is(1));
}

Answer 2

您可以使用这样的查询方法。使用下划线(_)得到属性child.

@Repository
public interface CustomerRepository extends JpaRepository<Customer, Long> {
    List<Customer> findByAddresses_City(String city);
}

Java Spring 具有@OneToMany 关系的数据@Query returns 无结果

Java Spring Data @Query with @OneToMany relation returns no result

java

spring

jpa

assertion