使用JPA

Question

我有2个表：User和Loan。 User有3个字段：id（PK），first_name和last_name。 Loan表的字段user_id是User表的外键：

通过保留新的Loan我需要创建新的User，如果他的first_name和last_name是唯一的，否则将id放入uder_id }}

我的贷款类的源代码：

 public class Loan {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;
    private Long sum;
    @ManyToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "user_id")
    private User user;

... methods ...

我正在使用此方法来保留新的User：

@PersistenceContext
private EntityManager em; 

public void save(User user) {
    if (user.getId() == null) {
         em.persist(user);
    } else {
         em.merge(user);
    }
}

当我尝试保存新的Loan时，它始终会保留一个新的User first_name和last_name但id不同。

 loan.setSum(sum);
 loan.setUser(new User(firstName, lastName));
 loanService.save(loan);

要使用用户的first_name和last_name作为PK不是解决方案，我需要id。

UPDATE_1

我试图通过他的名字找到User：

public User findByName(String firstName, String lastName) {
        TypedQuery<User> query = em.createQuery(
                "SELECT u FROM User u WHERE u.firstName = :firstName " +
                        "AND u.lastName = :lastName", User.class)
                .setParameter("firstName", firstName).setParameter("lastName", lastName);
        return query.getSingleResult();
    }

但是当我输入新用户时，我得到了一个例外：

javax.faces.FacesException: #{loanBean.requestLoan()}: javax.persistence.NoResultException: getSingleResult() did not retrieve any entities.

当我输入现有用户时，它会添加一个新的用户firstName和lastName，但新的id。 当我重复这个操作时，我得到了另一个例外：

javax.servlet.ServletException: javax.persistence.NonUniqueResultException: More than one result was returned from Query.getSingleResult()

UPDATE_2 非常感谢Pietro Boido提供了非常有用的建议。我在DB中的first_name和last_name字段上创建了唯一索引，并重构了save()方法。但是现在当我输入现有用户的数据时，我得到了新的异常

javax.servlet.ServletException: org.springframework.transaction.TransactionSystemException: Could not commit JPA transaction; nested exception is javax.persistence.RollbackException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.6.0.v20150309-bf26070): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLIntegrityConstraintViolationException: The statement was aborted because it would have caused a duplicate key value in a unique or primary key constraint or unique index identified by 'FIRST_LAST_NAME' defined on 'USER'.
Error Code: 20000

Answer 1

您应首先执行查询以按名称查找用户，并仅在未找到时创建新用户：

User user = userService.find(firstName, lastName);
if (user == null) {
    user = loanService.createUser(new User(firstName, lastName));
}
loan.setSum(sum);
loan.setUser(user);
loanService.save(loan);

由于可能没有给定名称的用户，因此在查询用户时请使用getResultList，因为getSingleResult总是希望找到结果。

List<User> users = query.getResultList();
if (!users.isEmpty()) {
    return users.iterator().next();
} else {
    return null;
}

代码假定数据库在first_name，last_name。

上有唯一索引

您不应该对ManyToOne关系进行级联操作。想一想：如果删除贷款，是否也应该删除用户？

当相关实体是相关实体的一部分并且它们的生命周期一起管理时，应该使用级联操作。

@Entity
@Table(name="loans")
public class Loan implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;
    @Column(name = "total")
    private Long sum;
    @ManyToOne()
    @JoinColumn(name = "user_id")
    private User user;

    //...
}

这是一个可能的工作示例：

@Stateless
public class LoanService implements LoanServiceRemote {

    @PersistenceContext
    private EntityManager em;

    @Override
    public User createUser(User user) {
        em.persist(user);
        return user;
    }

    @Override
    public Loan createLoan(Loan loan) {
        em.persist(loan);
        System.out.println("loan persisted: id=" + loan.getId());
        return loan;
    }

    @Override
    public Loan saveLoan(Loan loan) {
        em.merge(loan);
        return loan;
    }

    @Override
    public Long incrementLoan(Integer loanId, long amount) {
        Loan loan = em.find(Loan.class, loanId);
        if (loan != null) {
            long sum = loan.getSum() + amount;
            /*
             * The entity is bound to the entity manager,
             * because it was returned by the find method.
             * We can simply set its properties and
             * the entity manager will update the datasource
             * after the method returns and the transaction commits. 
             * No need to call persist or merge.
             */
            loan.setSum(sum);
            return sum;
        }
        return null;
    }

    @Override
    public boolean deleteLoan(Integer loanId) {
        Loan loan = em.find(Loan.class, loanId);
        if (loan != null) {
            em.remove(loan);
            return true;
        }
        return false;
    }

    @Override
    public Loan findLoan(Integer loanId) {
        return em.find(Loan.class, loanId);
    }

    @Override
    public List<Loan> requestLoans(LoanRequest loanRequest) {
        User user;
        TypedQuery<User> query = em.createQuery("select user from User user where user.firstName = :firstName and user.lastName = :lastName", User.class);
        query.setParameter("firstName", loanRequest.getFirstName());
        query.setParameter("lastName", loanRequest.getLastName());
        List<User> users = query.getResultList();
        if (users.isEmpty()) {
            user = new User();
            user.setFirstName(loanRequest.getFirstName());
            user.setLastName(loanRequest.getLastName());
            //new entities must be persisted
            em.persist(user);
        } else {
            user = users.get(0);
        }

        List<Loan> loans = new ArrayList<>();
        Long[] totals = loanRequest.getTotals();
        for (int i = 0; i < totals.length; i++) {
            Loan loan = new Loan();
            loan.setSum(totals[i]);
            loan.setUser(user);
            em.persist(loan);
            loans.add(loan);
        }

        return loans;
    }
}

单元测试：

@Test
public void testLoan() {
    User user = loanService.createUser(newUser());

    Loan loan1 = new Loan();
    loan1.setSum(10L);
    loan1.setUser(user);

    Loan loan2 = loanService.createLoan(loan1);
    assertNotNull(loan2);

    Integer loanId = loan2.getId();

    assertNotNull(loanId);
    assertEquals(loan1.getSum(), loan2.getSum());
    assertEquals(loan1.getUser(), user);

    User user2 = loanService.createUser(newUser());
    loan2.setUser(user2);
    loan2.setSum(20L);
    Loan loan3 = loanService.saveLoan(loan2);
    assertLoanEquals(loan2, loan3);

    Long total = loanService.incrementLoan(loanId, 10L);
    assertNotNull(total);
    assertEquals((Long)(loan3.getSum() + 10L), total);

    loan3.setSum(total);

    Loan loan4 = loanService.findLoan(loanId);
    assertLoanEquals(loan3, loan4);

    boolean result = loanService.deleteLoan(loanId);
    assertTrue(result);

    Loan loan5 = loanService.findLoan(loanId);
    assertNull(loan5);

    Long[] totals = new Long[]{1L,2L,3L};
    LoanRequest loanRequest = new LoanRequest();
    loanRequest.setFirstName("Amerigo");
    loanRequest.setLastName("Vespucci");
    loanRequest.setTotals(totals);

    List<Loan> loans = loanService.requestLoans(loanRequest);
    assertNotNull(loans);
    assertEquals(3, loans.size());
    for (int i = 0; i < 3; i++) {
        assertEquals(totals[i], loans.get(i).getSum());
        loanService.deleteLoan(loans.get(i).getId());
    }


}

void assertLoanEquals(Loan loan1, Loan loan2) {
    assertNotNull(loan1);
    assertNotNull(loan2);
    assertEquals(loan1.getSum(), loan2.getSum());
    assertUserEquals(loan1.getUser(), loan2.getUser());
    assertEquals(loan1.getId(), loan2.getId());
}
void assertUserEquals(User user, User user2) {
    assertNotNull(user);
    assertNotNull(user2);
    assertEquals(user.getId(), user2.getId());
    assertEquals(user.getFirstName(), user2.getFirstName());
    assertEquals(user.getLastName(), user2.getLastName());
}