In my application I have three entities, BaseNotification as a parent and SmsNotification and EmailNotification as child entities which extend BaseNotification.
I am using hibernate single table inheritance strategy so all attributes fit into one table.
Lets say I have some common attributes on BaseNotification entity, and child entities have one specific attribute each (SmsNotification has mobileNumber and EmailNotification has email attribute).
public abstract class BaseNotification implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(name = "name")
private String name;
#Column(name = "description")
private String description;
...
}
public abstract class EmailNotification extends BaseNotification implements Serializable {
private static final long serialVersionUID = 1L;
#Column(name = "email")
private String email;
...
}
public abstract class SmsNotification extends BaseNotification implements Serializable {
private static final long serialVersionUID = 1L;
#Column(name = "mobileNumber")
private String mobileNumber;
...
}
Is there a way of fetching all notifications with all attributes while sorting on child attributes (for example by email) in one go with spring rest?
Related
I am new to Hibernate and JPA. I have several entities, each of which contains following four columns:
1. created_by
2. last_modified_by
3. created_date
4. last_modified_date
I would like these columns to get auto-populated while saving the associated entity.
Two sample entities are as follows:
Entity 1:
#Entity
#Table(name = "my_entity1")
#Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class MyEntity1 implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(name = "name")
private String name;
#Column(name = "created_by")
private String createdBy;
#Column(name = "last_modified_by")
private String lastModifiedBy;
#Column(name = "created_date")
private Instant createdDate;
#Column(name = "last_modified_date")
private String lastModifiedDate;
}
Entity 2:
#Entity
#Table(name = "my_entity2")
#Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class MyEntity2 implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(name = "description")
private String description;
#Column(name = "created_by")
private String createdBy;
#Column(name = "last_modified_by")
private String lastModifiedBy;
#Column(name = "created_date")
private Instant createdDate;
#Column(name = "last_modified_date")
private String lastModifiedDate;
}
In this context, I have gone through following posts: How to autogenerate created or modified timestamp field?, How can you make a created_at column generate the creation date-time automatically like an ID automatically gets created?.
I am getting how to capture the dates fields but I cannot understand how to capture created_by and last_modified_by.
Auditing Author using AuditorAware and Spring Security...
To tell JPA about currently logged in user we will need to provide an
implementation of AuditorAware and override getCurrentAuditor()
method. And inside getCurrentAuditor() we will need to fetch currently
logged in user.
Like this:
public class AuditorAwareImpl implements AuditorAware<String> {
#Override
public String getCurrentAuditor() {
return "TestUser";
// Can use Spring Security to return currently logged in user
// return ((User) SecurityContextHolder.getContext().getAuthentication().getPrincipal()).getUsername()
}
}
Now enable jpa auditing by using #EnableJpaAuditing
#Configuration
#EnableJpaAuditing(auditorAwareRef = "auditorAware")
public class JpaConfig {
#Bean
public AuditorAware<String> auditorAware() {
return new AuditorAwareImpl();
}
}
Look at this to get more details....
I'm trying to update my user entity and I have an error that comes to mind:
ERROR: A NULL value violates the NOT NULL constraint of the "id" column Detail: The failed row contains (null, 1, 1)
The problem surely stems from my relationship between user and profile which is n-n
public class Utilisateur implements Serializable {
private static final long serialVersionUID = 1L;
#Id
private Integer id;
private Integer fixe;
private Boolean deleted;
private Boolean actif;
private String email;
private Integer mobile;
private String motDePasse;
private String nom;
private String prenom;
#ManyToMany
private List<Profil> profils = new ArrayList<Profil>();
public Utilisateur() {
}
}
public class Profil implements Serializable {
private static final long serialVersionUID = 1L;
#Id
private Integer id;
private String codeProfil;
private String libelleProfil;
#JsonManagedReference
#ManyToMany
private List<MenuAction> menuActions = new ArrayList<MenuAction>();
public Profil() {
}
}
How you generate value for your id?
Seems you need some way to generate value for you ID.
For example, use #GeneratedValue, like:
#GeneratedValue(strategy = IDENTITY)
I am new in Spring Data Rest. I want to save a parent entity with his children. The class are Distribution and FileIdVersion.
This is the Distribution entity.
#Entity
#DistributionValidator
public class Distribution extends AbstractAuditableJpaEntityImpl {
private static final long serialVersionUID = 1L;
#NotNull
#Length(min = 1, max = 256)
#SafeHtml
private String company;
#OneToMany(cascade = CascadeType.ALL, mappedBy = "distribution")
#Size(max = 256)
private List<FileIdVersion> fileIdVersions = new ArrayList<>();
public Distribution() {
super();
}
public Distribution(final String company, final String name, final String topic, final ZonedDateTime uploadDate,
final ZonedDateTime setupDate, final UUID uuid, final List<FileIdVersion> fileIdVersions,
final List<Bundle> bundles, final List<String> recipientId) {
super();
this.company = company;
this.fileIdVersions = fileIdVersions;
}
}
This is the FileIdVersion entity.
#Entity(name = "bundle_file_id_version")
public class FileIdVersion extends AbstractJpaEntityImpl implements Serializable {
private static final long serialVersionUID = 1L;
#NotNull
#FileId
private String fileId;
#FileVersion
private String fileVersion;
#ManyToOne
#NotNull
#JsonIgnore
private Bundle bundle;
public FileIdVersion() {}
}
I want to save one distribution object with his fileIdVersion. I am trying something like this:
This request only persist in BBDD one record of distribution, but not persist any records in FileIdVersion entity. How I can to persist the distribution with his file id versions? Thank you in advance!!!
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;
}
For example I want to make relationship between User A and User B and they have RelationshipEntity named MakeFriend, I am used code below, but I am also want to set in relation entity some property values like role = 10.........
userRepository.createRelationshipBetween(startUser, endUser, MakeFriend.class, RelTypes.FRIEND.name());
#RelationshipEntity
public class MakeFriend {
#GraphId
private Long id;
private String role;
#StartNode
private UserEntity startUser;
#EndNode
private UserEntity endUser
#NodeEntity
public class UserEntity implements Serializable {
private static final long serialVersionUID = 1L;
public static final String FRIEND = "FRIEND";
public static final String JOYNED = "JOYNED";
#GraphId
private Long id;
#Indexed(unique = true)
private Long userId;
private String email;
You could could add the following to your UserEntity class:
#RelatedToVia(type = RelTypes.FRIEND, direction = Direction.BOTH)
private MakeFriend friend;
friend.setRole("yourRole");
Another way to do it, when you're using the advanced mapping mode, is using one of the NodeBacked.relateTo() methods. Then add the property to the returned Relationship.
And a third way, it to use the Neo4jTemplate.createRelationshipBetween() method and provide your properties (e.g. role) as the final argument.