How to implement a custom Hibernate enum type indexed on a label

LabeledEnum.java

public interface LabeledEnum {
    String getLabel();
}

LabeledEnumType.java

import org.hibernate.HibernateException;
import org.hibernate.engine.spi.SessionImplementor;
import org.hibernate.usertype.DynamicParameterizedType;
import org.hibernate.usertype.UserType;

import java.io.Serializable;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Types;
import java.util.Properties;

public final class LabeledEnumType implements DynamicParameterizedType, UserType {

    private Class<? extends Enum> enumClass;

    @Override
    public Object assemble(Serializable cached, Object owner)
            throws HibernateException {
        return cached;
    }

    @Override
    public Object deepCopy(Object value) throws HibernateException {
        return value;
    }

    @Override
    public Serializable disassemble(Object value) throws HibernateException {
        return (Serializable) value;
    }

    @Override
    public boolean equals(Object x, Object y) throws HibernateException {
        return x == y;
    }

    @Override
    public int hashCode(Object x) throws HibernateException {
        return x == null ? 0 : x.hashCode();
    }

    @Override
    public boolean isMutable() {
        return false;
    }

    @Override
    public Object nullSafeGet(ResultSet rs, String[] names, SessionImplementor session, Object owner)
            throws HibernateException, SQLException {
        String label = rs.getString(names[0]);
        if (rs.wasNull()) {
            return null;
        }
        for (Enum value : returnedClass().getEnumConstants()) {
            if (value instanceof LabeledEnum) {
                LabeledEnum labeledEnum = (LabeledEnum) value;
                if (labeledEnum.getLabel().equals(label)) {
                    return value;
                }
            }
        }
        throw new IllegalStateException("Unknown " + returnedClass().getSimpleName() + " label");
    }

    @Override
    public void nullSafeSet(PreparedStatement st, Object value, int index, SessionImplementor session)
            throws HibernateException, SQLException {
        if (value == null) {
            st.setNull(index, Types.VARCHAR);
        } else {
            st.setString(index, ((LabeledEnum) value).getLabel());
        }
    }

    @Override
    public Object replace(Object original, Object target, Object owner)
            throws HibernateException {
        return original;
    }

    @Override
    public Class<? extends Enum> returnedClass() {
        return enumClass;
    }

    @Override
    public int[] sqlTypes() {
        return new int[]{Types.VARCHAR};
    }

    @Override
    public void setParameterValues(Properties parameters) {
        ParameterType params = (ParameterType) parameters.get( PARAMETER_TYPE );
        enumClass = params.getReturnedClass();
    }
}

Role.java

public enum Role implements LabeledEnum {
    ADMIN("admin"), USER("user"), ANONYMOUS("anon");

    private final String label;

    Role(String label) {
        this.label = label;
    }

    @Override
    public String getLabel() {
        return label;
    }
}

User.java

public class User {

    @Id
    private int id;

    @Type(type = "com.example.hibernate.LabeledEnumType")
    private Role role;
}

Thanks Jon!
I've been looking for a solution to mapping enums in a fancy way and this one is perfect because is reusable!

why not use
@javax.persistence.Enumerated(EnumType.STRING) ?

public class User {

    @Id
    private int id;

    @javax.persistence.Enumerated(EnumType.STRING)
    private Role role;
}

@ErwanLeroux it was 5 years ago I created this gist. I’m not using Hibernate or ORM anymore. I guess support for enums is much better now.

why not use
@javax.persistence.Enumerated(EnumType.STRING) ?

The presented code allows to store some String that's different to the enum's name. Which is still cool, should you want that.