In my Spring/JPA/Hibernate/Envers PostgreSQL application I'm trying to implement Spring Data Auditing
I have a following entity:
#Audited
#AuditTable("levels_history")
#Entity
#Table(name = "levels")
public class Level extends BaseEntity implements Serializable {
private static final long serialVersionUID = 642499791438799548L;
#Id
#SequenceGenerator(name = "levels_id_seq", sequenceName = "levels_id_seq", allocationSize = 1)
#GeneratedValue(strategy = GenerationType.AUTO, generator = "levels_id_seq")
private Long id;
private String name;
#NotAudited
#OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "level")
private List<Card> card = new ArrayList<Card>();
#CreatedBy
#Column(name = "created_by_user_id")
private User createdByUser;
#LastModifiedBy
#Column(name = "last_modified_by_user_id")
private User lastModifiedByUser;
where I have added:
#CreatedBy
#Column(name = "created_by_user_id")
private User createdByUser;
#LastModifiedBy
#Column(name = "last_modified_by_user_id")
private User lastModifiedByUser;
at the database level I want to write INTEGER ID for created_by_user_id and last_modified_by_user_id but during execution my application throws an error that created_by_user_id should be bytea type.
What am I doing wrong and how to configure Spring Data Auditing to use INTEGER for this purpose ?
Related
I have two table user(id,name) and user_mails(id,email,user_id) user to user_mails have one to many relation.
I have created following entity in spring boot
User
#Entity
#Table(name = "user")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id")
private int id;
#Column(name = "name", nullable = false)
private String name;
#OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
private Set<UserMail> userMails =new HashSet<UserMail>(0);
//Getter setter and constructor
}
UserMail
#Entity
#Table(name = "user_mails")
public class UserMail {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int id;
#Column(name = "email", nullable = false)
private String name;
#ManyToOne(fetch = FetchType.LAZY, optional = false)
#JoinColumn(name = "user_id")
private User user;
It is showing following output on calling controller
[{"id":1,"name":"Ram","userMails":[{"id":2,"name":"ram#b.com","user":{"id":1,"name":"Ram","userMails":[{"id":2,"name":"ram#b.com","user":{"id":1,"name":"Ram","userMails":[{"id":2,"name":"ram#b.com","user":{"id":1,"name":"Ram","userMails":[{"id":2,"name":"ram#b.com","user":{"id":1,"name":"Ram","userMails":[{"id":2,"name":"ram#b.com","user":{"id":1,"name":"Ram","userMails":[{"id":2,"name":"ram#b.com","user":{"id":1,"name":"Ram","userMails":
and more
I want to access all users with all mail ids also want to acces mail id with user details
What changes should I do to get proper result
I have two entities in one to many relationship in my spring-data-jpa project.
Parent entity -
#Entity
#Table(name = "code_group")
public class CodeGroup implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.SEQUENCE)
private Long id;
#NaturalId
#Column(nullable = false)
private String entityId;
#OneToMany
#JoinColumn(name = "codeGroupId", referencedColumnName = "entityId")
private List<SystemCode> systemCodes;
// .. getters setters
}
Child entity -
#Entity
#Table(name = "system_code")
public class SystemCode implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.SEQUENCE)
private Long id;
#Column(nullable = false)
private String codeGroupId;
#Column(nullable = false)
#Enumerated(EnumType.STRING)
private ActiveOrInactive status;
// getters and setters
}
status column is of enum type, it can only have Active or Inactive value.
My existing code works fine. It is fetching code group with associated system codes. I want to filter system code with status='Active'. How to do this?
Try like this:
#OneToMany
#JoinColumn(name = "codeGroupId", referencedColumnName = "entityId")
#Where(clause = "status= 'Active'")
private List<SystemCode> systemCodes;
You can create such methods in your repo:
List<CodeGroup> getAllBySystemCodes_Status(ActiveOrInactive status);
default List<CodeGroup> getAllActive() {
return getAllBySystemCodes_Status(ActiveOrInactive.Active);
}
default List<CodeGroup> getAllInactive() {
return getAllBySystemCodes_Status(ActiveOrInactive.Inactive);
}
I'm starting to learn Spring Java Framework . I created some Enity to join 2 Model like my Database. And now I want to insert to Join Table by JpaRepository. What i have to do?
This is my Code (Please fix help me me if something is not right)
Model Users_RoomId to define Composite Primary Key
#Embeddable
public class Users_RoomId implements Serializable {
private static final long serialVersionUID = 1L;
#Column(name = "ID_room", nullable = false)
private String idRoom;
#Column(name = "user_id", nullable = false)
private int idUser;
}
Model Users_Room to join 2 Model Users and Room
#Entity
#Table(name ="bookroom")
public class Users_Room {
#EmbeddedId
private Users_RoomId usersroomId;
#ManyToOne
#MapsId("idRoom")
private Room room;
#ManyToOne
#MapsId("idUser")
private Users users;
#Column(name = "Bookday")
private String bookday;
Model Users and Room I used annotation #OneToMany
Model Users
#Entity
#Table(name = "users")
public class Users implements Serializable{
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "user_id", nullable = false)
private int id;
#Column(name = "name", nullable = false)
private String name;
#Column(name = "email")
private String email;
#Column(name = "pass")
private String pass;
#Column(name = "role")
private int role;
#OneToMany(mappedBy = "users")
private List<Users_Room> user;
Model Room
#Entity
#Table(name ="room")
public class Room implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "ID_room", nullable = false)
private String id;
#Column(name = "name_room", nullable = false)
private String name;
#Column(name = "Description")
private String describe;
#ManyToOne
#JoinColumn(name = "ID_status")
private Status status;
#Column(name = "room_image")
private String image;
public Room() {
super();
}
#ManyToOne
#JoinColumn(name = "ID_kind")
private KindRoom kind;
#OneToMany(mappedBy = "room")
private List<Users_Room> rooms;
This is my database
So I don't know how to insert a new bookroom with iduser,idroom and bookday with JPA repository.. It'necessary to write Query in JPARepository or We just need to use method save() to insert data
Thanks everyone
I had same problem and solved with following code. I used method save() to insert data. Following code is 'createRoom' method in 'RoomService.java'.
RoomService.java
private final RoomRepository roomRepository;
private final UserRoomRepository userRoomRepository;
private final UserRepository userRepository;
public RoomService(RoomRepository roomRepository, UserRoomRepository userRoomRepository, UserRepository userRepository) {
this.roomRepository = roomRepository;
this.userRoomRepository = userRoomRepository;
this.userRepository = userRepository;
}
#Transactional
public RoomDto createRoom(Long userId, Long chattingUserId) {
Room room = roomRepository.save(new Room());
room.addUserRoom(userRepository.findById(userId).orElseThrow(()->new NoSuchElementException("No User")));
room.addUserRoom(userRepository.findById(chattingUserId).orElseThrow(()->new NoSuchElementException("No User")));
userRoomRepository.save(new UserRoom(userRepository.findById(userId).orElseThrow(()->new NoSuchElementException("No User")),room));
userRoomRepository.save(new UserRoom(userRepository.findById(chattingUserId).orElseThrow(()->new NoSuchElementException("No User")),room));
RoomDto roomDto = RoomDto.of(room);
return roomDto;
}
I have an unidirectional OneToMany JPA entity mapping in my (Spring Framework + Spring Data + Hibernate JPA) project. Entity classes are like in the following code.(I have removed irrelevant class members for brevity).
#Entity
#Table(name = "employees")
class Employee{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Integer id;
#OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
#JoinColumn(name = "employee_id")
private List<DepartmentAssignment> departmentAssignments = new ArrayList<>();
}
#Entity
#Table(name = "department_assignments")
class DepartmentAssignment{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Integer id;
#NotNull
#Column(name = "employee_id")
private Integer employeeId;
#NotNull
#Column(name = "department_id")
private Integer departmentId;
}
#Entity
#Table(name = "departments")
class Department{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Integer id;
}
And, in one of my service classes have a method to remove a DepartmentAssignment from an Employee like below.
public Employee deleteDepartmentAssignment(Integer empId, Integer deptAssignmentId) {
Employee employee = employeeRepository.findOne(empId);
if(employee != null) {
for ( DepartmentAssignment da : employee.getDepartmentAssignments()) {
if(da.getId().equals(deptAssignmentId)) {
employee.getDepartmentAssignments().remove(da);
employee = employeeRepository.save(employee);
break;
}
}
}
return employee;
}
However, calling above methods gives me an error: org.hibernate.exception.ConstraintViolationException ,and in the SQL log, I can see Column 'employee_id' cannot be null error for the last SQL statement of the transaction.
Can anybody tell me what I'm doing wrong here and how to get it fixed?
You don't need to add
#NotNull
#Column(name = "employee_id")
private Integer employeeId;
to the Employee, if you use #JoinColumn(name = "employee_id"). Try to remove it.
You can try the following, not sure why you use the plain id in the object. Thats not object relational mapping.
For more details see Hibernate triggering constraint violations using orphanRemoval
#Entity
#Table(name = "employees")
class Employee{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#OneToMany(cascade = CascadeType.ALL, mappedBy = "employee", orphanRemoval = true)
private List<DepartmentAssignment> departmentAssignments = new ArrayList<>();
}
#Entity
#Table(name = "department_assignments")
class DepartmentAssignment{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#ManyToOne(optional=false)
private Employee employee;
#ManyToOne(optional=false)
private Department department;
}
#Entity
#Table(name = "departments")
class Department{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
}
You must look .hbm.xml file and you should mapping your Entity in this file and
you can look this example
http://www.mkyong.com/hibernate/hibernate-one-to-many-relationship-example/
I hope it will be useful for you.
try removing
cascade = CascadeType.ALL
but im not 100% sure..
I have two JPA entities: User and Message.
Each Message has one sender and one receiver of type User. And on the other side each User has two sets of type Message: inbox and outbox.
Message:
#Entity
#Table(name = "messages")
public class Message {
#Id
#GeneratedValue(strategy = GenerationType.AUTO, generator = "seq")
#SequenceGenerator(name = "seq", sequenceName = "MESSAGES_SEQ")
#Column(name = "ID", insertable = false, updatable = false)
private int id;
#ManyToOne
#JoinColumn(name = "SENDER")
private User sender;
#ManyToOne
#JoinColumn(name = "RECEIVER")
private User receiver;
private String subject, content;
private Date sdate;
//getters and setters
}
All the properties which not being mapped with an annotation has he same name as the columns in database and are automatically mapped by JPA.
User:
#Entity
#Table(name = "users")
public class User {
#Column(name = "USERNAME")
private String username;
#Column(name = "PASSWORD")
private String pass;
#Column(name = "EMAIL")
private String email;
#Id
#GeneratedValue(strategy = GenerationType.AUTO, generator = "seq")
#SequenceGenerator(name = "seq", sequenceName = "USERS_SEQ")
#Column(name = "ID", insertable = false, updatable = false)
private int id;
#OneToMany(mappedBy = "uploader")
private Set<Book> books;
#OneToMany(mappedBy = "receiver")
private Set<Message> inbox;
#OneToMany(mappedBy = "sender")
private Set<Message> outbox;
//getters and setters
}
The problem is, when I select an User from Oracle database, then the inbox property is empty. How is this caused and how can I solve it?