java-具有规范,分页和条件fetch-join的Spring Data JPA存储库
作者:互联网
我正在使用具有规范和分页功能的Spring Data JPA存储库为实体列表实现搜索/过滤服务.我正在尝试减少查询数量(n 1个问题)并使用标准获取机制来获取嵌套数据.
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...
}
以及搜索/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
有办法解决这个问题吗?
findAll(DelegationFilter filter)(无分页的方法)的工作原理类似于魅力…仅使用join(无获取)也可以正常工作(即使有分页)
我知道JPQL有解决方案:
Spring-Data FETCH JOIN with Paging is not working
但我想坚持使用标准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
解决方法:
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,以便如果它是Long(计数查询返回的类型),则可以加入,否则进行获取.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,但这是我在ATM上遇到的唯一解决方案.
标签:spring-data-jpa,spring-data,hibernate,spring,java 来源: https://codeday.me/bug/20191118/2025978.html