one-to-many mapping using a foreign key, how?

theBernd · **Joined:** Wed Mar 16, 2005 10:18 am **Posts:** 4

Hi all,

we're using Hibernate 2.x and I have a simple problem which I don't know how to solve (even after reading this forum and the specs). Two tables:

[code]

create table list_something (
id int primary key,
translation_id int foreign key references translation(translation_id)
...,
);

create table translation (
id int primary key,
translation_id int not null,
locale_id int not null,
name varchar(100),
...
[/code]

Please note that list_something.translation_id (not the PK!) references translation_id in translation (which isn't a PK either and not unique; translation_id plus locale_id gives an alternate key, but...).

What I'd like to achive is getting a Set of translation objects for the referenced translation_id.

Any ideas how to solve this?

TIA! - Bernd

rascaly2k · **Joined:** Wed Feb 23, 2005 10:26 am **Posts:** 22

check this out
http://www.hibernate.org/hib_docs/v3/re ... n-onetoone

jklaps · **Posted:** Tue Jun 28, 2005 5:30 am

theBernd, did you manage to find a solution for your translation problem? I'm interested to read it...

Thanks.

theBernd · **Joined:** Wed Mar 16, 2005 10:18 am **Posts:** 4

Hi, no, I didn't understand the suggestion above and since we have been under extreme time pressure we made a datamodel change (hack would be more appropriate).

I read the release notes to Hibernate 3.x (comparision to 2.x, new features, whatever) and it sounds to me like this is a new 3.x feature.

fdelisle · **Posted:** Tue Jun 28, 2005 7:09 am

Hi theBernd,

First of all,

Quote:

references translation_id in translation which isn't a PK either and not unique

Well, if translation_id in translation isn`t unique, you shouldn't using a set. A set doesn't allows duplicate elements.

Second,

Quote:

translation_id plus locale_id gives an alternate key, but...)

I would like to know what is the end of your sentence here, but here is what I would suggest. Use a composite-key for your set and override .equals() and .hashcode in the translation object to reflect the translation table natural key (translation_id + locale_id) :

Code:

public class Translation implement Serializable
{
    private Serializable id; // i prefer to use Serializable for ids because this let me change its implementation later without altering this code
    private Serializable translationId;
    private Serializable localeId; // you should consider using a java.util.Locale object...

    [...]

    public boolean equals( Object o ) 
    {
        if( this == o ) return true;
        if( !(o instanceof Translation) ) return false;

        Translation other = (Translation)o;

        return   getTranslationId().equals( other.getTranslationId() )
                    && getLocaleId().equals( other.getLocaleId() );
    }

    public int hashCode()
    {
        int result;

        result = getTranslationId();
        result = 29 * result + getLocaleId();

        return result;
    }
}

Then, map your set in the list_something object this way:

Code:

<set name="translations" table="translation"> <!-- use cascade, inverse, etc as necessary-->

    <key>
        <column name="translation_id"/>
   <column name="locale_id"/>
    </key>

    <one-to-many
        class="Translation"/>
    
</set>

Let me know about it,

theBernd · **Joined:** Wed Mar 16, 2005 10:18 am **Posts:** 4

Hi Frank,

thanks for the info.

The related work has been my first Hibernate project and at the same time I had to find my way through another framework I haven't work with before. Since even co-workers with quite some Hibernate experience didn't know how to solve this, we changed the datamodel to make things work.

I really appreciate your help and will look back at it ASAP (month end means a two projects deadline). I can't remember what I meant with the ",but...". I think it was something along the lines that hibernate just cares about / references the primary key and not alternate keys, especially not composite ones (which may be wrong).

Anyway, thanks a lot. - Bernd