Spring Data JPA存储库，具有规范、分页和标准fetch-join

Question

@Entity
@Table(name = "delegations")
public class Delegation {

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

    @ManyToOne
    private Customer customer;

    // more fields, getters, setters, business logic...

}

@Entity
@Table(name = "customers")
public class Customer {

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

    // more fields, getters, setters, business logic...
}

DTO筛选器类：

public class DelegationFilter {

    private String customerName;

    // more filters, getters, setters...
}

和搜索/筛选服务：

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：

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

我使用的是Spring Boot 1.4（Spring 4.3.2,spring-data-jpa 1.10.2）和Hibernate 5.0.9

Answer 1

我也面临着同样的问题，我找到了一个解决办法（来源）。

您可以在运行时检查查询的返回类型，这样，如果它是long（计数查询返回的类型），则联接，否则获取。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()));
    };
}
...

我知道这不是很干净，但这是我唯一的解决办法。