我无法使用 JpaRepository 获得我想要的东西。我将尝试使用以下代码来解释我想要的内容:
存储库
@Repository
public interface CompanyRepository extends JpaRepository<Company, Long> {
Optional<Company> findByIdAndBranches_parent_idIsNull(Long id);
}
服务
@Service
public class BillService {
@Autowired
private CompanyRepository companyRepository;
@Autowired
private BranchRepository branchRepository;
public Company getCompanyById(Long id)
{
Optional<Company> company = companyRepository.findByIdAndBranches_parent_idIsNull(id);
return company.get();
}
公司实体类
public class Company
{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String name;
private String address;
@OneToMany(mappedBy = "company")
private List<Branch> branches;
@OneToMany(mappedBy = "company")
private List<User> users;
}
分支实体类
@Entity
public class Branch
{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String name;
private String address;
@JsonIgnore
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "company_id")
private Company company;
@JsonIgnore
@ManyToOne(cascade = CascadeType.ALL )
@JoinColumn(name = "parent_id")
@NotFound(action = NotFoundAction.IGNORE)
private Branch parent;
@OneToMany(mappedBy = "parent")
@NotFound(action = NotFoundAction.IGNORE)
private List<Branch> subBranches;
@OneToMany(mappedBy = "branch")
private List<User> users;
}
现在我想要的是公司的分支机构的父 ID 为 NULL,但无论父 ID 为何,我都会获取所有分支机构
这就是我想要的结果
{
"id": 1,
"name": "Lakshya",
"address": "Bahadurgarh",
"branches": [
{
"id": 1,
"name": "Lakshya Branch1",
"address": "Bahadurgarh1",
"subBranches": [
{
"id": 3,
"name": "Lakshya Branch1_3",
"address": "Bahadurgarh1_3",
"subBranches": [],
"users": []
}
],
"users": [
{
"id": 3,
"name": "User3_Company1_Branch1",
"address": "Bgz"
}
]
}
],
"users": [
{
"id": 1,
"name": "User_Company1",
"address": "Bgz"
},
{
"id": 2,
"name": "User_Company1_Branch1",
"address": "Bgz"
},
{
"id": 3,
"name": "User3_Company1_Branch1",
"address": "Bgz"
}
]
}
但这就是我得到的
{
"id": 1,
"name": "Lakshya",
"address": "Bahadurgarh",
"branches": [
{
"id": 1,
"name": "Lakshya Branch1",
"address": "Bahadurgarh1",
"subBranches": [
{
"id": 3,
"name": "Lakshya Branch1_3",
"address": "Bahadurgarh1_3",
"subBranches": [],
"users": []
}
],
"users": [
{
"id": 3,
"name": "User3_Company1_Branch1",
"address": "Bgz"
}
]
},
{
"id": 3,
"name": "Lakshya Branch1_3",
"address": "Bahadurgarh1_3",
"subBranches": [],
"users": []
}
],
"users": [
{
"id": 1,
"name": "User_Company1",
"address": "Bgz"
},
{
"id": 2,
"name": "User_Company1_Branch1",
"address": "Bgz"
},
{
"id": 3,
"name": "User3_Company1_Branch1",
"address": "Bgz"
}
]
}
最佳答案
如果需要过滤掉关联实体,可以使用@Where 。它不是 JPA 规范的一部分,但 Hibernate 的 JPA 实现提供了此注释。因此您可以指定任何附加条件:
public class Company
{
...
@OneToMany(mappedBy = "company")
@Where(clause = "parent_id is null")
private List<Branch> branches;
@OneToMany(mappedBy = "company")
private List<User> users;
}
然后您必须更改存储库的方法名称,因为您不再需要额外的条件:
@Repository
public interface CompanyRepository extends JpaRepository<Company, Long> {
Optional<Company> findById(Long id);
}
使用这种方法,您总是会得到只有分支机构而没有母公司的公司。如果您需要获取公司的所有分支机构(无论它们是否有父级),您可以为 Branch
实体类创建和使用存储库。
关于java - 如何在 Spring boot 中检查嵌套对象来获取对象?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56856775/