How do you map a “Map” in hibernate using annotati

Question

Using annotations how do you map a field in an entity which is a "Map" (Hashtable) of String to a given object? The object is annotated and instances of it are already stored in the hibernate databse.

I've found the syntax for definging a map with a simple key and value as such:

<class name="Foo" table="foo">
    ...
    <map role="ages">
         <key column="id"/>
         <index column="name" type="string"/>
         <element column="age" type="string"/>
     </map>
 </class>

And oddly with an entity as the key and a simple type as the value like so:

<class name="Foo" table="foo">
    ...
  <map role="ages">
    <key column="id"/>
    <index-many-to-many column="person_id" 
         class="Person"/>
    <element column="age" type="string"/>
  </map>
</class>
<class name="Person" table="person">
    ...
    <property name="name" column="name" 
         type="string"/>
</class>

But I don't see how to do this for a simple key to element mapping, and I don't see how to do this using annotations.

Answer 1

@CollectionOfElements(fetch = FetchType.LAZY)
@JoinTable(name = "JOINTABLE_NAME",
    joinColumns = @JoinColumn(name = "id"))
@MapKey(columns = @Column(name = "name"))
@Column(name = "age")
private Map<String, String> ages = new HashMap<String, String>();

Answer 2

I know this question is very old but maybe this could help someone.

Other posibility is something like that:

@Entity
@Table(name = "PREFERENCE", uniqueConstraints = { @UniqueConstraint(columnNames = { "ID_DOMAIN", "ID_USER", "KEY" })})
public class Preferences {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "ID", unique = true, nullable = false)
    private Long id;

    @Column(name = "ID_DOMAIN", unique = false, nullable = false")
    private Long domainId;

    @Column (name = "PREFERENCE_KEY")
    @Enumerated(EnumType.STRING)
    private PreferenceKey key;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "ID_USER", referencedColumnName = "ID")
    private User user;
}

and 

@Entity
@Table(name = "USER", uniqueConstraints = { @UniqueConstraint(columnNames = { "ID_DOMAIN", "LOGIN" })})
public class User {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "ID", unique = true, nullable = false)
    private Long id;

    @Column(name = "ID_DOMAIN", unique = false, nullable = false")
    private Long domainId;

    // more fields

    @ElementCollection(fetch = FetchType.LAZY)
    @JoinColumns({@JoinColumn(name = "ID_USER", referencedColumnName = "ID"), @JoinColumn(name = "ID_DOMAIN", referencedColumnName = "ID_DOMAIN")})
    @OneToMany(targetEntity = Preferences.class, fetch = FetchType.LAZY)
    @MapKey(name = "key")
    private Map<PreferenceKey, Preferences> preferencesMap;
}

That only produces two tables User and Preferences, note that PreferenceKey is unique for a User into a domain

Answer 3

You should probably use a UserType or UserCollectionType. Or, you can use a custom tupleizer.

see hibernate core documentation for the concepts and hibernate annotations documentation for the equivalent annotation approach.

Let me know if that isn't what you are asking for.

Answer 4

You could simply use the JPA annotation @MapKey (note that the JPA annotation is different from the Hibernate one, the Hibernate @MapKey maps a database column holding the map key, while the JPA's annotation maps the property to be used as the map's key).

@javax.persistence.OneToMany(cascade = CascadeType.ALL)
@javax.persistence.MapKey(name = "name")
private Map<String, Person> nameToPerson = new HashMap<String, Person>();