Question

Possible Duplicate:
DISTINCT() and ORDERBY issue

listOrders.DataSource = (from sp in dbdata.Specifications.Where(i => i.Approve == "Yes" && i.Awailable_BOM == "Yes")
                                     join ord in dbdata.Orders on sp.O_id equals ord.O_id
                                     join prd in dbdata.Products.Where(k => k.Hours_prd == null) on ord.O_id equals prd.O_ID
                                     orderby ord.Special, sp.Due_date
                                     select ord.O_id).Distinct();

I am tring to order by “ord.Special, sp.Due_date” but it doesn’t work after I put “Distinct();”. So how do I order by “ord.Special, sp.Due_date” and get distinct values.

Answer 1

I think this will return ordered ids:

listOrders.DataSource = 
   from sp in dbdata.Specifications.Where(i => i.Approve == "Yes" && i.Awailable_BOM == "Yes")
   join ord in dbdata.Orders on sp.O_id equals ord.O_id
   join prd in dbdata.Products.Where(k => k.Hours_prd == null) on ord.O_id equals prd.O_ID
   select new { ord.Special, sp.Due_date, ord.O_id } into x
   group x by x.O_id into g
   select new { g.Key, X = g.OrderBy(i => i.Special).ThenBy(i => i.Due_date).First()) } into y
   orderby y.X.Special, y.X.Due_date
   select y.Key;

Comments:

• select only fields you are interested in into x
• group those fields by field you want to be distinct into group g
• from each group select key (ord.O_id) and group item X, which will define order of id in result (in this case First occurrence of id will be returned)
• order results by X
• select Key (ord.O_id)