Hibernate : for a single table group attributes into a subclass
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I want to do :
create table user (
id bigint primary key auto_increment,
firstname varchar(128),
lastname varchar(128),
street varchar(128),
zipcode mediumint,
city varchar(128)
)
then :
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
private Address address; // Here I want an Address rather than street, zip, and city directly inside User
...
}
and :
public class Address {
private String street;
private Integer zipcode;
private String city;
...
}
(that is an example)
How to do that with Hibernate ? @OneToOne ? The problem seems simple and since I can't find the solution around here, it must be obvious.
java hibernate relational-database
asked Jan 4 at 12:04
JulienJulien
9531130
add a comment |
I want to do :
create table user (
id bigint primary key auto_increment,
firstname varchar(128),
lastname varchar(128),
street varchar(128),
zipcode mediumint,
city varchar(128)
)
then :
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
private Address address; // Here I want an Address rather than street, zip, and city directly inside User
...
}
and :
public class Address {
private String street;
private Integer zipcode;
private String city;
...
}
(that is an example)
How to do that with Hibernate ? @OneToOne ? The problem seems simple and since I can't find the solution around here, it must be obvious.
java hibernate relational-database
asked Jan 4 at 12:04
JulienJulien
9531130
Look at embeddable type: stackoverflow.com/questions/19341838/…
– Sodala
Jan 4 at 12:07
add a comment |
I want to do :
create table user (
id bigint primary key auto_increment,
firstname varchar(128),
lastname varchar(128),
street varchar(128),
zipcode mediumint,
city varchar(128)
)
then :
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
private Address address; // Here I want an Address rather than street, zip, and city directly inside User
...
}
and :
public class Address {
private String street;
private Integer zipcode;
private String city;
...
}
(that is an example)
How to do that with Hibernate ? @OneToOne ? The problem seems simple and since I can't find the solution around here, it must be obvious.
java hibernate relational-database
asked Jan 4 at 12:04
JulienJulien
9531130
I want to do :
create table user (
id bigint primary key auto_increment,
firstname varchar(128),
lastname varchar(128),
street varchar(128),
zipcode mediumint,
city varchar(128)
)
then :
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
private Address address; // Here I want an Address rather than street, zip, and city directly inside User
...
}
and :
public class Address {
private String street;
private Integer zipcode;
private String city;
...
}
(that is an example)
How to do that with Hibernate ? @OneToOne ? The problem seems simple and since I can't find the solution around here, it must be obvious.
java hibernate relational-database
java hibernate relational-database
asked Jan 4 at 12:04
JulienJulien
9531130
asked Jan 4 at 12:04
JulienJulien
9531130
asked Jan 4 at 12:04
JulienJulien
9531130
asked Jan 4 at 12:04
JulienJulien
9531130
asked Jan 4 at 12:04
JulienJulien
9531130
9531130
Look at embeddable type: stackoverflow.com/questions/19341838/…
– Sodala
Jan 4 at 12:07
add a comment |
Look at embeddable type: stackoverflow.com/questions/19341838/…
– Sodala
Jan 4 at 12:07
Look at embeddable type: stackoverflow.com/questions/19341838/…
– Sodala
Jan 4 at 12:07
Look at embeddable type: stackoverflow.com/questions/19341838/…
– Sodala
Jan 4 at 12:07
add a comment |
2 Answers
2
active
oldest
votes
First, annotate your Address class with @Embeddable:
@Embeddable
public class Address {
/*class definition here*/
...
}
And then update your user class in this way:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
@Embedded
private Address address;
...
}
That's it. If you want a more complex example look here. Just pay attention to model classes definition, it makes no difference if you are using Spring Boot or not as long as you handle your database with JPA.
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
1
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
add a comment |
You can associate entities through a one-to-one relationship using @OneToOne annotation.
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private Address address;
@OneToOne(cascade = CascadeType.ALL)
@PrimaryKeyJoinColumn
public Address getAddress()
{
return address;
}
}
Reference:
https://docs.jboss.org/hibernate/annotations/3.5/reference/en/html/entity.html#entity-mapping-association
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, annotate your Address class with @Embeddable:
@Embeddable
public class Address {
/*class definition here*/
...
}
And then update your user class in this way:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
@Embedded
private Address address;
...
}
That's it. If you want a more complex example look here. Just pay attention to model classes definition, it makes no difference if you are using Spring Boot or not as long as you handle your database with JPA.
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
1
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
add a comment |
First, annotate your Address class with @Embeddable:
@Embeddable
public class Address {
/*class definition here*/
...
}
And then update your user class in this way:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
@Embedded
private Address address;
...
}
That's it. If you want a more complex example look here. Just pay attention to model classes definition, it makes no difference if you are using Spring Boot or not as long as you handle your database with JPA.
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
1
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
add a comment |
First, annotate your Address class with @Embeddable:
@Embeddable
public class Address {
/*class definition here*/
...
}
And then update your user class in this way:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
@Embedded
private Address address;
...
}
That's it. If you want a more complex example look here. Just pay attention to model classes definition, it makes no difference if you are using Spring Boot or not as long as you handle your database with JPA.
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
First, annotate your Address class with @Embeddable:
@Embeddable
public class Address {
/*class definition here*/
...
}
And then update your user class in this way:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String firstname;
private String lastname;
@Embedded
private Address address;
...
}
That's it. If you want a more complex example look here. Just pay attention to model classes definition, it makes no difference if you are using Spring Boot or not as long as you handle your database with JPA.
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
answered Jan 4 at 12:12
LoreloreloreLorelorelore
2,30261829
2,30261829
1
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
add a comment |
1
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
1
1
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
Thanks it seems great. I will accept it when I can test it since some work needs to be done before that.
– Julien
Jan 4 at 12:17
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
@Julien you're welcome!
– Lorelorelore
Jan 8 at 15:39
add a comment |
You can associate entities through a one-to-one relationship using @OneToOne annotation.
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private Address address;
@OneToOne(cascade = CascadeType.ALL)
@PrimaryKeyJoinColumn
public Address getAddress()
{
return address;
}
}
Reference:
https://docs.jboss.org/hibernate/annotations/3.5/reference/en/html/entity.html#entity-mapping-association
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
add a comment |
You can associate entities through a one-to-one relationship using @OneToOne annotation.
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private Address address;
@OneToOne(cascade = CascadeType.ALL)
@PrimaryKeyJoinColumn
public Address getAddress()
{
return address;
}
}
Reference:
https://docs.jboss.org/hibernate/annotations/3.5/reference/en/html/entity.html#entity-mapping-association
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
add a comment |
You can associate entities through a one-to-one relationship using @OneToOne annotation.
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private Address address;
@OneToOne(cascade = CascadeType.ALL)
@PrimaryKeyJoinColumn
public Address getAddress()
{
return address;
}
}
Reference:
https://docs.jboss.org/hibernate/annotations/3.5/reference/en/html/entity.html#entity-mapping-association
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
You can associate entities through a one-to-one relationship using @OneToOne annotation.
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private Address address;
@OneToOne(cascade = CascadeType.ALL)
@PrimaryKeyJoinColumn
public Address getAddress()
{
return address;
}
}
Reference:
https://docs.jboss.org/hibernate/annotations/3.5/reference/en/html/entity.html#entity-mapping-association
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
answered Jan 4 at 12:11
Dinesh AcharyaDinesh Acharya
653
653
add a comment |
add a comment |
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Look at embeddable type: stackoverflow.com/questions/19341838/…
– Sodala
Jan 4 at 12:07