linq - How to get maximum Id in List and then return list -

i have structure this:

no  id  name status 1    1       1 2    1    b    1 3    1    c    1 4    1    d    1 5    2    e    3 6    2    f    3 7    2    g    3 

i want run linq when list results maximum row each status , row details well.like:

no  id  name status 4    1    d    1 7    2    g    3 

means latest entry status.

is there way around, have tried max, orderby descending single result need list result.

you have extract groups of same id (groupby), , export max no each group, selectmany :

public void exec() {     var items = new list<item>{          new item{ no = 1, id = 1, name = "a", status = 1} ,         new item{ no = 2, id = 1, name = "b", status = 1} ,         new item{ no = 3, id = 1, name = "c", status = 1} ,         new item{ no = 4, id = 1, name = "d", status = 1} ,         new item{ no = 5, id = 2, name = "e", status = 1} ,         new item{ no = 6, id = 2, name = "f", status = 1} ,         new item{ no = 7, id = 2, name = "g", status = 1} ,      };      var result = items                     .groupby(groupeditems => groupeditems.id)                     .selectmany(i => items                                          .where(inneritem => inneritem.id == i.key && inneritem.no == i.max(ii => ii.no))                                          .select(inneritem => inneritem)                                                                                       );      foreach (var item in result)         console.writeline("max item of id {0} : no = {1}, name = {2}, status = {3}", item.id, item.no, item.name, item.status);   }  private class item {     public int32 no { get; set; }     public int32 id { get; set; }     public string name { get; set; }     public int32 status { get; set; } } 

output :

max item of id 1 : no = 4, name = d, status = 1 max item of id 2 : no = 7, name = g, status = 1 

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alternative:

items.groupby(groupeditems => groupeditems.id)      .select(g => g.orderbydescending(x => x.no).first())