一个客户可以有多个帐户。一个帐户只属于一个客户。一个客户可以没有帐户，但一个帐户总是有一个客户。
我找不到如何在spring Boot 3.1.5（我想）中使用JPA来实现这一点。
它创建了正确的表，并插入工程，但阅读帐户的客户永远不会找到客户的帐户。
我使用的是H2 DB，它会在每次重启时自动生成模式（这是开发阶段所需的方法）
下面是客户实体：
@Data
@Entity
@NoArgsConstructor
public class Customer extends BaseEntity {
@Id
@SequenceGenerator(name= "customer_seq", sequenceName = "customer_seq", allocationSize = 1, initialValue = 100)
@GeneratedValue(strategy = GenerationType.SEQUENCE,generator = "customer_seq")
private Long id;
@Column(length = 256, nullable = false, unique = false)
private String email;
@NotNull(message = "Password cannot be null")
@Size(min = 8, max = 16, message = "Password must be between 8 and 16 characters long")
private String password;
// Where do you tell it the table to map to (i.e. Account) and the FK column in that table (i.e. customer_id)?
// If it can "guess" the table by the "Account" object name, why cant it guess the FK column name is "customer_id"?
@OneToMany(mappedBy = "customer", fetch = FetchType.EAGER)
private Set<Account> accounts;
public Customer(String email, String password) {
this.email = email;
this.password = password;
}
@Temporal(TemporalType.DATE) // not needed any more?
private Date dob; // or use sql.Date?
// other fields omitted for brevity.
}字符串
下面是我尝试过的各种组合的帐户实体：
@Data
@Entity
@NoArgsConstructor
public class Account {
@Id
@SequenceGenerator(name= "account_seq", sequenceName = "account_seq", allocationSize = 1, initialValue = 100)
@GeneratedValue(strategy = GenerationType.SEQUENCE,generator = "account_seq")
private Long id;
BigDecimal balance = new BigDecimal(0);
// How can you tell it that "customer" is the FK to the ID of the Customer Table? It should be able to guess that.
@ManyToOne(optional = false)
// runs, but doesnt find any accounts for any customer.
// @OneToOne(optional = false)
// runs, but doesnt find any accounts for any customer.
// @JoinColumn
// This one runs, but never finds accounts for any customer.
// @JoinColumn(name = "customer_id", nullable = false)
// Fails with: Column 'id' is duplicated in mapping for entity 'com.ohds.bomj.entity.Account'
// @JoinColumn(name = "id", nullable = false)
// This one also runs, but never finds accounts for any customer.
@JoinColumn(name = "customer_id", referencedColumnName = "id", nullable = false) // doesnt run
// This one fails with: references a column named 'customer_id' but the target entity 'com.ohds.bomj.entity.Customer' has no property which maps to this column
// @JoinColumn(name = "customer_id", referencedColumnName = "customer_id", nullable = false)
Customer customer;
public Account(Customer c) {
this.customer = c;
}
}型
如果您查看H2控制台的以下屏幕截图，我们可以看到它在Account上创建了预期的“CUSTOMER_ID”列，并正确地使用ID为100的客户填充了该列。
然而，FK看起来不像它指向customer.id FK应该指向客户表中的ID。
的数据
下面我们可以看到customer表中的客户，其中沿着一个ID为100的客户。
的
要尝试读取客户帐户，我使用以下控制器：
@RestController
class CustomerController {
Logger log = LoggerFactory.getLogger(CustomerController.class);
private final CustomerRepository repository;
CustomerController(CustomerRepository repository) {
this.repository = repository;
}
@GetMapping("/customers")
List<Customer> all() {
List<Customer> customers = repository.findAll();
// for each customer, get the accounts
for (Customer customer : customers) {
log.info("customer: " + customer);
log.info("accounts: " + customer.getAccounts());
}
return customers;
}
}型
输出如下所示：
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : customer: Customer(id=1, firstName=null, lastName=null, [email protected], password=haha123, phone=null, accounts=[], dob=null)
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : accounts: []
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : customer: Customer(id=2, firstName=null, lastName=null, [email protected], password=foobarr123, phone=null, accounts=[], dob=null)
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : accounts: []
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : customer: Customer(id=100, firstName=bob, lastName=null, email=haha@blabla, password=12345678, phone=null, accounts=[], dob=1969-07-05)
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : accounts: []
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : customer: Customer(id=101, firstName=null, lastName=null, email=ano[email protected], password=12345678, phone=null, accounts=[], dob=null)
2023-11-13T23:32:28.770+01:00 INFO 19908 --- [nio-8085-exec-1] c.o.bomj.controller.CustomerController : accounts: []型
正如我们所看到的，它未能找到客户100的帐户。
仓库看起来像这样：
public interface CustomerRepository extends JpaRepository<Customer, Long> {}
and
public interface AccountRepository extends JpaRepository<Account, Long> {}型
在Grails中，两个简单的注解将创建所需的FK并自动进行连接。
2条答案
按热度按时间dfuffjeb1#
问题是，我猜，因为你说插入工作，当你持久化数据时，你实际上并没有将你的Account示例添加到Customer.accounts集合中。这些是POJO，JPA维护这些对象的缓存，所以如果你不维护关系的两端并保持它与数据库同步，你会得到一个空的集合，直到你刷新对象或重新启动应用程序。你不认为后者是一个解决方案，因为你一直在重新创建表。
您可以通过在其中一个示例上尝试entityManager.refresh（account）并验证调用后的状态是否与数据库中的预期状态匹配来证明这一点。
要解决这个问题而不强制查询数据库，只需确保将Account添加到Customer，并使Account引用Customer。
bkhjykvo2#
@Chris是对的，这是lombok造成的问题。似乎你不能使用**@Data**，尽管我已经看到使用它的例子。然而，@Getter和@Setter似乎是好的。
当lombok @Data被删除时，我得到了无限递归，即使是懒惰加载，因为返回一个Customer对象包括Accounts，其中包括Customers，其中包括Accounts等。为了解决这个问题，我在Account中的Customer字段中添加了**@JsonIgnore**。
我还没有检查@ToString是否能与JPA一起工作。我不知道我是否需要实现hashcode或equals。Equals有点无实际意义，因为所有客户帐户都是唯一的（即不能有两个相同的电子邮件）。
下面是工作代码（与原始问题不同的地方）
账户.java
字符串
客户端.java
型
输出到控制台
型
复位端点输出
型