【发布时间】:2021-02-26 15:03:09
【问题描述】:
我正在尝试同时使用两个 Hibernate 功能:EntityGraph 和 Criteria API TypedQuery,但无法解决如何一起使用它们:
@Override
public List<Customer> findByFilterEntityGraph(final CustomerFilter filter) {
final EntityManager entityManager = sessionFactory.createEntityManager();
final CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
final CriteriaQuery<Customer> criteriaQuery = criteriaBuilder.createQuery(Customer.class);
final Root<Customer> customerMetamodel = criteriaQuery.from(Customer.class);
final Predicate name = criteriaBuilder.like(customerMetamodel.get("name"), filter.getName());
final Predicate age = criteriaBuilder.between(
customerMetamodel.get("age"), filter.getAgeFrom(), filter.getAgeTo());
final Predicate predicate = criteriaBuilder.and(name, age);
criteriaQuery.where(predicate);
final TypedQuery<Customer> query = entityManager.createQuery(criteriaQuery);
final EntityGraph<?> entityGraph = entityManager.getEntityGraph("customer.products");
Map<String, Object> properties = Map.of("javax.persistence.fetchgraph", entityGraph);
// ... ?
return null;
}
我确定我的EntityGraph 声明正确,因为它在其他地方也可以使用。但我找不到如何使用它们。
我知道这样的方法:
public Optional<Customer> findByIdEntityGraph(final long customerId) {
final EntityManager entityManager = sessionFactory.createEntityManager();
final EntityGraph<?> entityGraph = entityManager.getEntityGraph("customer.products");
Map<String, Object> properties = Map.of("javax.persistence.fetchgraph", entityGraph);
final Customer customer = entityManager.find(Customer.class, customerId, properties);
return Optional.ofNullable(customer);
}
允许将 Criteria API 与 EntityGraph 一起使用,但对于单一结果且没有条件,如使用 TypedQuery 的过滤器。
列表可以同时使用EntityGraph和TypedQuery吗?
【问题讨论】:
标签: java hibernate hibernate-criteria criteria-api criteriaquery