我正在为实体列表实现搜索/过滤服务,使用具有规范和分页功能的 Spring Data JPA 存储库。我正在尝试减少查询数量(n+1 问题)并使用条件获取机制获取嵌套数据。
我有两个实体类:
@Entity
@Table(name = "delegations")
public class Delegation {
@Id
@GeneratedValue(strategy = IDENTITY)
private Long id;
@ManyToOne
private Customer customer;
// more fields, getters, setters, business logic...
}
和
@Entity
@Table(name = "customers")
public class Customer {
@Id
@GeneratedValue(strategy = IDENTITY)
private Long id;
// more fields, getters, setters, business logic...
}
DTO 过滤器类:
public class DelegationFilter {
private String customerName;
// more filters, getters, setters...
}
和搜索/过滤服务:
public class DelegationService {
public Page<Delegation> findAll(DelegationFilter filter, Pageable page) {
Specifications<Delegation> spec = Specifications.where(
customerLike(filter.getCustomerName())
);
return delegationRepository.findAll(spec, page);
}
public List<Delegation> findAll(DelegationFilter filter) {
Specifications<Delegation> spec = Specifications.where(
customerLike(filter.getCustomerName())
);
return delegationRepository.findAll(spec);
}
private Specification<Delegation> customerLike(String customerName) {
return (root, query, cb) -> {
Join<Delegation,Customer> join = (Join) root.fetch(Delegation_.customer);
return cb.like(cb.lower(join.get(Customer_.name)), addWildCards(customerName.toLowerCase()));
};
}
private static String addWildCards(String param) {
return '%' + param + '%';
}
}
问题:
当我调用 findAll(DelegationFilter filter, Pageable page)
时出现异常:
org.springframework.dao.InvalidDataAccessApiUsageException:
org.hibernate.QueryException: query specified join fetching, but the owner
of the fetched association was not present in the select list
有没有办法解决这个问题?
findAll(DelegationFilter filter)
(没有分页的方法)像魅力一样工作......仅使用 join
(没有 fetch
)也可以正常工作(即使有分页)
我知道有针对 JPQL 的解决方案: Spring-Data FETCH JOIN with Paging is not working 但我想坚持标准 api...
我正在使用 Spring Boot 1.4(spring 4.3.2、spring-data-jpa 1.10.2)和 Hibernate 5.0.9
最佳答案
我遇到了同样的问题,并且找到了解决方法 ( source )。
您可以在运行时检查查询的返回类型,如果它是 Long
(计数查询返回的类型),您就加入,否则就获取。在您的代码中,它将如下所示:
...
private Specification<Delegation> customerLike(String customerName) {
return (root, query, cb) -> {
if (query.getResultType() != Long.class && query.getResultType() != long.class) {
Join<Delegation,Customer> join = (Join) root.fetch(Delegation_.customer);
} else {
Join<Delegation,Customer> join = root.join(Delegation_.customer);
}
return cb.like(cb.lower(join.get(Customer_.name)), addWildCards(customerName.toLowerCase()));
};
}
...
我知道它不是很干净,但这是我从 ATM 上找到的唯一解决方案。
关于java - 具有规范、分页和标准获取连接的 Spring Data JPA 存储库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39614628/