Hibernate @ManyToOne and @OneToMany with Liquibase

user_6912711 0
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I have been on this for serveral nights. I would like to link number in user (one user to many numbers) with user in number (many numbers to one user). I have no luck and need your knowledge. No matter what I do I always get errors with either this or that. A straight answer of what to do will do.

application.properties:

spring.jpa.hibernate.ddl-auto=validate

User entity:

@Entity
@Table(name = "user")
public class Users implements Serializable {

  private static final long serialVersionUID = 2323232323L;

  @Id
  @Column(nullable = false)
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  private Long id;

  @OneToMany(cascade = CascadeType.ALL, mappedBy = "user")
  private List<Number> number;

Number entity:

@Entity
@Table(name = "number")
public class Number implements Serializable {

  private static final long serialVersionUID = 1212121212L;

  @Id
  @Column(nullable = false)
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  private Long id;

  @ManyToOne(cascade = CascadeType.ALL)
  private Users user;

Liquibase:

<createTable tableName="user">
      <column name="id" type="BIGINT(8)" autoIncrement="true">
        <constraints nullable="false" primaryKey="true"/>
      </column>
      <column name="fk_number" type="BIGINT"/>
    </createTable>

    <createTable tableName="number">
      <column name="id" type="BIGINT(8)" autoIncrement="true">
        <constraints nullable="false" primaryKey="true"/>
      </column>
      <column name="user" type="BIGINT"/>
    </createTable>
liquibase spring-boot jpa hibernate java
2 Answers

There are 2 issues you need to fix:

  1. In a relational table model, you normally model a many-to-one/one-to-many association with a foreign key column on the many side. So, in your example, you only need a foreign key column on the number table and not on the user table.

  2. If you don't specify a @JoinColumn, Hibernate expects that the name of the foreign key column follows this pattern <name of the attribute that owns association>_<name of the primary key of referenced entity>. In your example, Hibernate expects a number_id column in the number table. You can learn more about association mappings in my Ultimate Guide - Association Mappings with JPA and Hibernate.

It should work, if you keep your entity mappings and use this table definition:

<createTable tableName="user">
  <column name="id" type="BIGINT(8)" autoIncrement="true">
    <constraints nullable="false" primaryKey="true"/>
  </column>
</createTable>

<createTable tableName="number">
  <column name="id" type="BIGINT(8)" autoIncrement="true">
    <constraints nullable="false" primaryKey="true"/>
  </column>
  <column name="user_id" type="BIGINT"/>
</createTable>
user_89943850

As per my understanding, you will be having a foreign key in number table which will represent the user to whom the number will be associated.

Since you are not mentioning @JoinColumn (the annotatiuon which holds information about foreign key)while specifying the @ManyToOne relationship in your Number Entity, thus, by default jpa is trying to look for the column named user_id in your number table(which is not available)

Just add the annotation @JoinColumn giving proper attributes , it should work.

@ManyToOne(cascade = CascadeType.ALL)
@JoinColumn(name="<<name of foreign key in number table>>", refrencedColumnName = "id")

refrencedColumnName tells which column has to referred in the parent entity.

user_76718760
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