CODEWITHSUNDEEP
hibernatejpaspring-boot
Ankit Bansal
Ankit Bansal

Reputation: 2358

Hibernate Entities Definition

I am new to hibernate. I am designing an application in spring boot with JPA Hibernate.

I have database schema as : 

CREATE TABLE IF NOT EXISTS `offers` (
`id` int(11) NOT NULL,
  `detail` varchar(255) NOT NULL
 `status` varchar(255) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

CREATE TABLE `restaurants` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `name` varchar(255) NOT NULL,
 `status` varchar(255) NOT NULL
 PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1

CREATE TABLE `rest_offer_map` (
 `id` int(11) NOT NULL,
 `rest_id` int(11) NOT NULL,
 `offer_id` int(11) NOT NULL
 `status` varchar(255) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1

I guess schema is clear. Now It could be the case that an restaurant exists without any offer. How do I create entities so that I could achieve left join. Also there is a column named status in every table, it needs to be taken care of (especially in left join as in that case status will be a join condition)

The o/p I want is list of all restaurants with the offers(in any), so left join is needed.

Upvotes: 0

Views: 96

Answers (1)

Justus Ikara
Justus Ikara

Reputation: 172

You need to create the Restaurant entity as the owning side of the relationship as below 

@Entity
@Table(name="restaurants")
public class Restaurant implements Serializable{
@Id
@Column(name="id")
private Long id;
@ManyToMany
@JoinTable(name="rest_offer_map",joinColumns=@JoinColumn(name="rest_id", referencedColumnName="id"),              inverseJoinColumns=@JoinColumn(name="offer_id",referencedColumnName="id"))
private Collection<Offer> offers;
        .....

}

And the Offer entity as below

@Entity
@Table(name"offers")
public class Offer implements Serializable{
     @Id
     @Column(name="id")
     private Long id;
     ManyToMany(mappedBy="offers")
     private Collection<Restaurant> restaurants;
......
}

Upvotes: 2

Related Questions