You can use the overload that takes an instance:
map.Type(new XXXUserType(5));
RP
On Thursday, October 23, 2014 11:33:59 AM UTC+1, pasquale wrote:
>
> In the past, I have had the requirement to create an implementation of
> IUserType interface to manage multiples columns of a table as an array
> (10 columns of type int).
> Now I have the same requirement: 5 colums of type int.
> I would like to add a constructor to IUserType implementation to handle
> both case (5 columns and 10 columns).
> It's possible to achieve this goal via mapping by code using the
> declaration of Type<TPersistentType>(object parameters)? I can't figure
> out how I can accomplish my task.
>
> Property(x => x.XXX, map => {
> map.Columns(cm => cm.Name("xxx1"),
> cm => cm.Name("xxx2"),
> ...);
> map.Type<XXXUserType>(???);
> map.Access(Accessor.NoSetter);
> });
>
> Thanks in advance.
>
>
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