java – JPA CriteriaBuilder – 按一对多关系中的关联实体数排序

我有两个实体Customer和Order一对多关系.
对于每个客户,我需要计算相关订单的数量,并按此数字对结果进行排序.
在本机postgres查询中,它看起来像这样：

select cust.id, count(order.id) from customers cust
left outer join orders order
on cust.id = order.customer_id
where .... conditions ...
group by cust.id
order by count desc;

但我必须使用CriteriaBuilder执行此操作,因为此查询是使用CriteriaBuilder添加其他条件的更大代码段的一部分.在Hibernate中我可能会使用Projections,但我在JPA中找不到类似的东西.

任何使用CriteraBuilder编写查询的帮助都将非常感激.

先感谢您.

解决方法:

假设实体Customer具有如下所示的OneToMany属性：

@OneToMany(mappedBy = "customerId")
private Collection<Order> orders;

您可以使用以下查询：

EntityManager em;  // to be built or injected
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Tuple> cq = cb.createTupleQuery();
Root<Customer> customer = cq.from(Customer.class);
CollectionJoin<Customer, Order> orders = customer.join(Customer_.orders, JoinType.LEFT);
cq.select(cb.tuple(customer, cb.count(orders)));
cq.where(... add some predicates here ...);
cq.groupBy(customer.get(Customer_.id));
cq.orderBy(cb.desc(cb.count(orders)));
List<Tuple> result = em.createQuery(cq).getResultList();
for (Tuple t : result) {
    Customer c = (Customer) t.get(0);
    Long cnt = (Long) t.get(1);
    System.out.println("Customer " + c.getName() + " has " + cnt + " orders");
}

上述方法使用Metamodel.如果您不喜欢它,可以将Customer_.orders替换为“orders”,将Customer_.id替换为“id”.

如果OneToMany属性属于另一种类型,请将CollectionJoin替换为正确类型的集合(ListJoin,SetJoin,MapJoin).

标签：criteria-api,java,jpa,hibernate
来源： https://codeday.me/bug/20190823/1696098.html