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c# - Usage of Server Side Controls in MVC Frame work -

i using asp.net 4.0 , mvc 2.0 web application. project requiremrnt have use server side control in application not possibl in noraml case. ideally want use adrotator control , datalist control. i saw few samples , references in codepleax mvc controllib howwver found less useful. can tell how utilize theese controls in asp.net application along mvc. note: please provide functionalities related adrotator , datalist controls not equivalent functionalities thanks in advace. mvc pages not use normal .net solution makes use of normal .net components impossible. a normal .net page use event driven solution call different methods service side mvc use actions , view completly different way handle things. also, mvc not use viewstate normal .net controlls require. found article discussing mixing of normal .net , mvc.

entity framework 4 - Ordering sub-collections in an EF/LINQ query -


i having trouble figuring out how (and where) order subcollection in linq/ef query.

basically have collection of subscriptiontypes, each having own collection of subscriptions. each subscription in collection ordered numberofmonths.

here current query:

    public iqueryable<subscriptiontype> subscriptiontypesbyproperty(string propertycode) {     return in db.subscriptiontypes.include("subscriptions")            a.subscriptions.any(x => x.propertycode == propertycode)            select a; } 

i subscriptions ordered numberofmonths. tried this:

    public iqueryable<subscriptiontype> subscriptiontypesbyproperty(string propertycode)     {         return in db.subscriptiontypes.include("subscriptions")                a.subscriptions.orderby(q => q.numberofmonths).any(x => x.propertycode == propertycode)                select a;     } 

.. did not order subscriptions correctly.

does know of easy way this?

update: subscriptiontype , subscription types generated ef designer.

thanks, adam

change select part sth

   select new subscriptiontype{ ?,?,?,                      subscriptions = a.subscription.orderby(b=>b.?)} 

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