Java Spring JPA关联实体创建相同的记录
我正在设置弹簧靴应用程序 尝试创建“添加用户菜单”，每次我发布JSON数据时，如以下示例所示： MySQL总是在Java Spring JPA关联实体创建相同的记录,java,json,spring,jpa,Java,Json,Spring,Jpa,我正在设置弹簧靴应用程序 尝试创建“添加用户菜单”，每次我发布JSON数据时，如以下示例所示： MySQL总是在role\u表中创建新的管理记录 我的目的是只在角色表中为用户表中的多个用户创建一条管理记录（一对一） 到目前为止，我已经尝试修改了几个SpringJPA注释，它们是不同类型的获取、关联和配置 到目前为止，我的MySql数据库表上的结果如下： role_table. user_table id. role. id username role_id 1 Ad
role\u表中创建新的管理记录
我的目的是只在角色表
中为用户表
中的多个用户创建一条管理记录（一对一）
到目前为止，我已经尝试修改了几个SpringJPA注释，它们是不同类型的获取、关联和配置
到目前为止，我的MySql数据库表上的结果如下：
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Admin 4. Roy. 5
5. Admin. 5. Subrey. 4
6. User. 6. Alhanus. 7
7. Guest 7. Rsidi. 6
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Roy. 1
5. Subrey. 3
6. Alhanus. 1
7. Rsidi. 1
public class Role implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="id", length=8, unique = true)
private int id;
@Column(name = "role",nullable = false, unique = false)
private String role;
@OneToMany(mappedBy="role", cascade=CascadeType.ALL, fetch=FetchType.LAZY)
@Fetch(value = FetchMode.SUBSELECT)
private List<User> user = new ArrayList<User>();}
public class User implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="user_id", length=8, unique = true)
private int user_id;
@Column(name = "username", nullable = false, unique = true)
private String username;
@Column(name = "password", nullable = false, unique = true)
private String password;
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
private Role role;}
public class AdminController {
private static final Logger LOGGER = LoggerFactory.getLogger(AdminController.class);
@Autowired
private UserRepository userRepository;
@Autowired
private RoleRepository roleRepository;
private Role role = new Role();
@Autowired
private BCryptPasswordEncoder passwordEncoder;
@PreAuthorize("hasAnyRole('ADMIN')")
@PostMapping("/admin/add")
public String addUserByAdmin(@RequestBody User user){
String pwd = user.getPassword();
String encryptedPwd = passwordEncoder.encode(pwd);
user.setPassword(encryptedPwd);
if(userRepository.findByUsername(user.getUsername())==null){
userRepository.save(user);
} else {
return "user already exist
}
return "user added successfully ... ";
}
我希望MySql表中的结果如下：
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Admin 4. Roy. 5
5. Admin. 5. Subrey. 4
6. User. 6. Alhanus. 7
7. Guest 7. Rsidi. 6
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Roy. 1
5. Subrey. 3
6. Alhanus. 1
7. Rsidi. 1
public class Role implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="id", length=8, unique = true)
private int id;
@Column(name = "role",nullable = false, unique = false)
private String role;
@OneToMany(mappedBy="role", cascade=CascadeType.ALL, fetch=FetchType.LAZY)
@Fetch(value = FetchMode.SUBSELECT)
private List<User> user = new ArrayList<User>();}
public class User implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="user_id", length=8, unique = true)
private int user_id;
@Column(name = "username", nullable = false, unique = true)
private String username;
@Column(name = "password", nullable = false, unique = true)
private String password;
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
private Role role;}
public class AdminController {
private static final Logger LOGGER = LoggerFactory.getLogger(AdminController.class);
@Autowired
private UserRepository userRepository;
@Autowired
private RoleRepository roleRepository;
private Role role = new Role();
@Autowired
private BCryptPasswordEncoder passwordEncoder;
@PreAuthorize("hasAnyRole('ADMIN')")
@PostMapping("/admin/add")
public String addUserByAdmin(@RequestBody User user){
String pwd = user.getPassword();
String encryptedPwd = passwordEncoder.encode(pwd);
user.setPassword(encryptedPwd);
if(userRepository.findByUsername(user.getUsername())==null){
userRepository.save(user);
} else {
return "user already exist
}
return "user added successfully ... ";
}
谢谢大家
像这样的角色类：
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Admin 4. Roy. 5
5. Admin. 5. Subrey. 4
6. User. 6. Alhanus. 7
7. Guest 7. Rsidi. 6
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Roy. 1
5. Subrey. 3
6. Alhanus. 1
7. Rsidi. 1
public class Role implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="id", length=8, unique = true)
private int id;
@Column(name = "role",nullable = false, unique = false)
private String role;
@OneToMany(mappedBy="role", cascade=CascadeType.ALL, fetch=FetchType.LAZY)
@Fetch(value = FetchMode.SUBSELECT)
private List<User> user = new ArrayList<User>();}
public class User implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="user_id", length=8, unique = true)
private int user_id;
@Column(name = "username", nullable = false, unique = true)
private String username;
@Column(name = "password", nullable = false, unique = true)
private String password;
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
private Role role;}
public class AdminController {
private static final Logger LOGGER = LoggerFactory.getLogger(AdminController.class);
@Autowired
private UserRepository userRepository;
@Autowired
private RoleRepository roleRepository;
private Role role = new Role();
@Autowired
private BCryptPasswordEncoder passwordEncoder;
@PreAuthorize("hasAnyRole('ADMIN')")
@PostMapping("/admin/add")
public String addUserByAdmin(@RequestBody User user){
String pwd = user.getPassword();
String encryptedPwd = passwordEncoder.encode(pwd);
user.setPassword(encryptedPwd);
if(userRepository.findByUsername(user.getUsername())==null){
userRepository.save(user);
} else {
return "user already exist
}
return "user added successfully ... ";
}
AdminController类，如下所示：
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Admin 4. Roy. 5
5. Admin. 5. Subrey. 4
6. User. 6. Alhanus. 7
7. Guest 7. Rsidi. 6
role_table. user_table
id. role. id username role_id
1 Admin 1 Tony 1
2. User 2. Buddy. 2
3. Guest. 3. Liaska. 3
4. Roy. 1
5. Subrey. 3
6. Alhanus. 1
7. Rsidi. 1
public class Role implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="id", length=8, unique = true)
private int id;
@Column(name = "role",nullable = false, unique = false)
private String role;
@OneToMany(mappedBy="role", cascade=CascadeType.ALL, fetch=FetchType.LAZY)
@Fetch(value = FetchMode.SUBSELECT)
private List<User> user = new ArrayList<User>();}
public class User implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name="user_id", length=8, unique = true)
private int user_id;
@Column(name = "username", nullable = false, unique = true)
private String username;
@Column(name = "password", nullable = false, unique = true)
private String password;
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
private Role role;}
public class AdminController {
private static final Logger LOGGER = LoggerFactory.getLogger(AdminController.class);
@Autowired
private UserRepository userRepository;
@Autowired
private RoleRepository roleRepository;
private Role role = new Role();
@Autowired
private BCryptPasswordEncoder passwordEncoder;
@PreAuthorize("hasAnyRole('ADMIN')")
@PostMapping("/admin/add")
public String addUserByAdmin(@RequestBody User user){
String pwd = user.getPassword();
String encryptedPwd = passwordEncoder.encode(pwd);
user.setPassword(encryptedPwd);
if(userRepository.findByUsername(user.getUsername())==null){
userRepository.save(user);
} else {
return "user already exist
}
return "user added successfully ... ";
}
问题在于初始化角色
对象的持久性。在这里，您将获得JSON中的用户对象，如下所示：
{"username": "Frantu", "password": "Pshycho8k8ss", "role" : {"role":"ADMIN"} }
没有为角色实体初始化id字段
因此，您的用户对象将是
User {id=0, username = Frantu, password= Pshycho8k8ss , role = {id=0, role=ADMIN }
您没有在此处传递role.id
，因此JPA将为每个新请求创建一个带有新主键的新记录
要解决此问题，首先，使用role
属性从数据库（如果存在）获取role
，并将其设置为用户对象
Role dbRole = roleRepository.findByRole(user.getRole().getRole());
if(dbRole!=null){
user.setRole(dbRole);
}
userRepository.save(user);
您如何保存角色。在AdminController中，我只能看到创建了一个新的角色对象。它在哪里与用户链接？你能添加代码的这一部分吗？我只是简单地用名为postman的应用程序发布JSON对象，spring处理一切，我不创建链接用户，spring引导自动处理，我发布的JSON数据看起来像这样“用户名”：“Frantu”，“password”：“Pshycho8k8ss”，“role”：{“role”：“ADMIN”}，因为您只是作为管理员传递角色，而没有主键。它被视为一个新记录，您在DB中有多个角色。我现在就知道了。。。谢谢大家。将cascadeType.ALL更改为cascadeType.MERGE工作…使用JSON数据，如用户{id=0，用户名=Frantu，密码=Pshycho8k8ss，角色={id=0，角色=ADMIN}用户{id=0，用户名=Frantu，密码=Pshycho8k8ss，角色={id=0，角色=ADMIN}对我来说真的工作…将cascadeType.ALL更改为cascadeType.MERGE…谢谢你的帮助。