Hm ok thats an idea and its working, however you have to be very careful for chained models.
SortableDataModel -> ListDataModel -> List<Typ>. Calling wrappedData on the first one is still not Serializable, you will have to call it twice to get the real Data. But thats implementation specific so its ok - thx for the tipp - tried to save the whole bean which failed, but saving the data only works. A method like "getRealWrappedData" would be nice, to get the real "leaf" of the DataModel Tree which maybe a chain, opinions? kind regards Am Montag, den 07.08.2006, 10:05 -0400 schrieb Mike Kienenberger: > On 8/7/06, Torsten Krah <[EMAIL PROTECTED]> wrote: > > Hm - thats not possible i guess. Using a SortableDataModel ( for > > example ) which is not serializable by default, you have to make a > > completly custom solution, extending and so on, to make it work. > > So this parameter to preserve the model is nice, you get a stateful > > model which can easily be sort by different SortCriterions. > > > > If you use sortable="true", your DataModel is automatically wrapped into > > a SortableDataModel and you won't be able to use saveState component, > > how did you solved this? > > > > kind regards > > > > Am Freitag, den 04.08.2006, 11:08 -0400 schrieb Mike Kienenberger: > > > For what it's worth, I never use preserveDataModel=true. Instead I > > > perform a t:saveState on my backing list. > > > > > Again, you don't use saveState on your model. You use it on the data > inside the model. However, I'm not sure if that's always possible. > Again, I've never tried it, but DataModel.[gs]etWrappedData should > return something you can serialize if you're not already making your > underlying data available via accessors.
