Spring Data JPA规范Manyto One Realization

Question

我对Spring Data JPA是新手，有两个实体，很少使用stackoverflow的例子，但运气不好。

用户-->*纳税

private static final long serialVersionUID = 1L;

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

纳税人。JVA

公共类纳税实现可序列化{

private static final long serialVersionUID = 1L;

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

@NotNull
@Column(name = "payment_date", nullable = false)
private LocalDate paymentDate;

@NotNull
@Column(name = "amount", precision=10, scale=2, nullable = false)
private BigDecimal amount;

@Column(name = "reference")
private String reference;

@ManyToOne
private User user;

public Long getId() {
    return id;
}..

}

private TaxPaymentSpecification(){

}

static Specification<TaxPayment> hasUser(Long userId){
    return new Specification<TaxPayment>() {

        @Override
        public Predicate toPredicate(Root<TaxPayment> arg0, CriteriaQuery<?> arg1,
                CriteriaBuilder arg2) {
            // TODO Auto-generated method stub
            Root<TaxPayment> root = arg0;
            Subquery<Long> subqry =  arg1.subquery(Long.class);
            Root<User> user =  subqry.from(User.class);

            final Join<User,TaxPayment> taxpays = root.join("user");

            subqry.select(taxpays.<Long> get("user_id"));

            subqry.where(arg2.equal(user.<Long> get("id"),userId));
            return arg2.in(arg0.get("user_id")).value(subqry);

       }
    };

}

Answer 1

如果您已经知道userId，则不需要使用subquery，该userId将是TaxPayment表中user_id的值：

@Override
public Predicate toPredicate(Root<TaxPayment> arg0, CriteriaQuery<?> arg1, 
        CriteriaBuilder arg2) {
  return arg2.equal(arg0.get("user_id"), userId);
}