在Spring中保存实体时，引用请求体中具有列的子实体。

huangapple
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英文:

Save entity in spring referencing child entity with a column in request body

问题

我有3个实体：User（用户）、Office（办公室）和Role（角色）。Role和Office已经存在于数据库中。

在SQL中，User实体使用其id引用Office，并使用其名称引用Role。SQL中的记录示例：

  1. Role表：
  2. 1 | USER
  4. Office表：
  5. 1 | Office 1
  7. User表：
  8. John | Doe | john.doe@mail.com | Password123 | 1 | USER
  10. 对于User表：
  11. 1 -> 引用Office
  12. USER -> 引用Role

Role实体：

  1. @Data
  2. @Entity
  3. @NoArgsConstructor
  4. @AllArgsConstructor
  5. @Table(name = "t_role")
  6. public class Role {
  8.     @Id
  9.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  10.     private Integer id;
  12.     @Column(name = "name")
  13.     private String name;
  14. }

Office实体：

  1. @Data
  2. @Entity
  3. @Table(name = "t_office")
  4. public class Office {
  6.     @Id
  7.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  8.     private Integer id;
  10.     @Column(name = "name")
  11.     private String name;
  12. }

User实体：

  1. @Data
  2. @Entity
  3. @Builder
  4. @Table(name = "t_user")
  5. public class User {
  7.     @Id
  8.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  9.     private Integer id;
  11.     @Column(name = "first_name")
  12.     private String firstName;
  14.     @Column(name = "last_name")
  15.     private String lastName;
  17.     @Column(name = "email")
  18.     private String email;
  20.     @Column(name = "password")
  21.     private String password;
  23.     @ManyToOne(fetch = FetchType.LAZY, optional = false)
  24.     @JoinColumn(name="role", referencedColumnName="name")
  25.     private Role role;
  27.     @ManyToOne(fetch = FetchType.LAZY, optional = false)
  28.     @JoinColumn(name = "office_id", referencedColumnName="id")
  29.     private Office office;
  30. }

问题是：如何在不在请求体中指定所有Office和Role对象的情况下保存User？只需使用Office的id和Role的名称？也许需要使用DTO或Deserializer？

提前感谢。

我需要一个请求体如下：

  1. {
  2.   "firstName": "John",
  3.   "lastName": "Doe",
  4.   "email": "john.doe@mail.com",
  5.   "password": "Password123",
  6.   "officeId": 1,
  7.   "role": "USER"
  8. }

而不是一个请求体如下：

  1. {
  2.   "firstName": "John",
  3.   "lastName": "Doe",
  4.   "email": "john.doe@mail.com",
  5.   "password": "Password123",
  6.   "office": {
  7.     "id": 1,
  8.     "name": "Office 1"
  9.   },
  10.   "role": {
  11.     "id": 1,
  12.     "name": "USER"
  13.   }
  14. }
英文:

I have 3 entities: User, Office and Role. The Role and Office already exist in database.

In SQL the User entity references Office with its id and references Role with its name. Example of a record in SQL:

  1. Role table:
  2. 1 | USER
  4. Office table:
  5. 1 | Office 1
  7. User table:
  8. John | Doe | john.doe@mail.com | Password123 | 1 | USER
  10. For user table:
  11. 1 -> reference to Office
  12. USER -> reference to Role

Role entity:

  1. @Data
  2. @Entity
  3. @NoArgsConstructor
  4. @AllArgsConstructor
  5. @Table(name = "t_role")
  6. public class Role {
  8.     @Id
  9.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  10.     private Integer id;
  12.     @Column(name = "name")
  13.     private String name;
  14. }

Office entity:

  1. @Data
  2. @Entity
  3. @Table(name = "t_office")
  4. public class Office {
  6.     @Id
  7.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  8.     private Integer id;
  10.     @Column(name = "name")
  11.     private String name;
  12. }

User entity:

  1. @Data
  2. @Entity
  3. @Builder
  4. @Table(name = "t_user")
  5. public class User {
  7.     @Id
  8.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  9.     private Integer id;
  11.     @Column(name = "first_name")
  12.     private String firstName;
  14.     @Column(name = "last_name")
  15.     private String lastName;
  17.     @Column(name = "email")
  18.     private String email;
  20.     @Column(name = "password")
  21.     private String password;
  23.     @ManyToOne(fetch = FetchType.LAZY, optional = false)
  24.     @JoinColumn(name="role", referencedColumnName="name")
  25.     private Role role;
  27.     @ManyToOne(fetch = FetchType.LAZY, optional = false)
  28.     @JoinColumn(name = "office_id", referencedColumnName="id")
  29.     private Office office;
  30. }

The question is: How can I achieve saving a User without specifying all the Office and Role object in request body, but just the id for Office and name for Role ? Maybe it is needed to use a DTO or a Deserializer ?
Thanks in advance.

I need a request body like this:

  1. {
  2.   "firstName": "John",
  3.   "lastName": "Doe",
  4.   "email": "john.doe@mail.com",
  5.   "password": "Password123",
  6.   "officeId": 1,
  7.   "role": "USER"
  8. }

Instead of a request body like this:

  1. {
  2.   "firstName": "John",
  3.   "lastName": "Doe",
  4.   "email": "john.doe@mail.com",
  5.   "password": "Password123",
  6.   "office": {
  7.     "id": 1,
  8.     "name": "Office 1"
  9.   },
  10.   "role": {
  11.     "id": 1,
  12.     "name": "USER"
  13.   }
  14. }

答案1

得分: 2

是的，您需要使用DTO类来处理请求体部分。

  1. public class UserDTO {
  2.     private String firstName;
  4.     private String lastName;
  6.     private String email;
  8.     private String password;
  10.     private String role;
  12.     private Integer officeId;
  13. }

然后从数据库中获取办公室和角色的数据，并将数据设置到用户对象中，然后保存。例如：

  1. User user = convertUserDTOtoUser(userDTO);
  2. Optional<Office> office = officeRepository.findById(userDTO.getOfficeId());
  3. Role role = roleRepository.findByName(userDTO.getRole());
  4. user.setOffice(office);
  5. user.setRole(role);
  6. userRepository.save(user);

请注意，这是Java代码示例，用于处理请求体部分并与数据库交互。

英文:

Yes, you need to use a DTO class for the request body

  1. public class UserDTO {
  2.     private String firstName;
  4.     private String lastName;
  6.     private String email;
  8.     private String password;
  10.     private String role;
  12.     private Integer officeId;
  13. }

and fetch the data from the database for office and role from database and set the data in the user and then save. Ex:

  1. User user = convertUserDTOtoUser(userDTO);
  2. Optional&lt;Office&gt; office = officeRepository.findById(userDTO.getOfficeId());
  3. Role role = roleRepository.findByName(userDTO.getRole());
  4. user.setOffice(office);
  5. user.setRole(role);
  6. userRepository.save(user);

huangapple
  • 本文由 发表于 2020年8月2日 15:00:13
  • 转载请务必保留本文链接：https://go.coder-hub.com/63213266.html
  • java
  • json
  • rest
  • spring-boot
  • spring-data-jpa
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