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Hibernate version:3

Mapping documents:

Code between sessionFactory.openSession() and session.close():

Full stack trace of any exception that occurs:

Name and version of the database you are using:MySQL 4.1.9

The generated SQL (show_sql=true):

Debug level Hibernate log excerpt:

For a <one-to-many> relationship between 2 Objects, Hibernate requires for Lists an index column in the table of the child elements.
As explained in the documentation : “For lists, this column contains sequential integers numbered from zero. Make sure that your index really starts from zero if you have to deal with legacy data”.

Does Hibernate3 implement a mechanism to automatically generate / determine the index value to be set into the database_table.
If yes, howtodo ?

Thanks a lot for any hints that could help.
Cheers,
Laurent

I don't think so. I just override the setList method

Thanks Cris !

But what kind of a strategy would you advise for index management in your setList() method ?

Do you know whether hibernate enables to get the last index value without previously instantiating the list as it is able to do for primary keys ? or a mean to get the size() of the list still w/o instantiating the list ?

Thanks for your time and any additional hint that could help my search ?

cheers,
Laurent

You don't need to manage the index column for ordered collections! Just add the column to the mapping--Hibernate will use 0 for the first element in your list, and so on.

Read the manual --> http://www.hibernate.org/hib_docs/v3/re ... ns-indexed

Actually, I had read the manual (a couple of times ;-) ! We did add the column to the mapping ! I does work with SELECT (List()) when the index column is propely set with data.

Dealing with saving (INSERT) objects, the index column is not incremented by hibernate (the column remains with a null value) ; then, the SELECT does not work anymore !