LINQ到与多骨料，总结和Max SQL

Question

注意更新时间：所有的任何和所有帮助 谢谢，对于那些精明的vb.net这里是自包含的例子...

Structure Stuff 
     public client as String 
     public BDate as DateTime 
     public KB as long 
     public sub new(client as string, BDate as DateTime, KB as Long) 
      Me.client = client 
      Me.BDate = BDate 
      Me.KB = KB 
     end sub 
    End Structure 
Sub Main 

Dim Stuffs as new List(Of Stuff) 
Stuffs.Add(new Stuff("dev0db05", "6/25/2012 7:00:35 AM",1)) 
Stuffs.Add(new Stuff("dev0db05", "6/25/2012 7:00:35 AM",1)) 
Stuffs.Add(new Stuff("dev0db05", "6/25/2012 7:00:35 AM",1)) 
Stuffs.Add(new Stuff("dev0db05", "6/26/2012 7:00:35 AM",2)) 
Stuffs.Add(new Stuff("dev0db05", "6/26/2012 7:00:35 AM",2)) 
Stuffs.Add(new Stuff("dev0db05", "6/26/2012 7:00:35 AM",2)) 
Stuffs.Add(new Stuff("dev0db05", "6/27/2012 7:00:35 AM",3)) 
Stuffs.Add(new Stuff("dev0db05", "6/27/2012 7:00:35 AM",3)) 
Stuffs.Add(new Stuff("dev0db05", "6/27/2012 7:00:35 AM",3)) 
Stuffs.Add(new Stuff("dev0db05", "6/28/2012 7:00:35 AM",4)) 
Stuffs.Add(new Stuff("dev0db05", "6/28/2012 7:00:35 AM",4)) 
Stuffs.Add(new Stuff("dev0db05", "6/28/2012 7:00:35 AM",4)) 

dim q = From c In Stuffs _ 
    where c.client= "dev0db05" _ 
    order by c.BDate _ 
    group c by c.Client,c.bdate into g = group _ 
    select ClientName = Client, BDate = DateTime.Parse(bdate).ToShortDateString, SumKB = g.sum(Function(p) p.kb) 

Output: 
ClientName BDate SumKB 
dev0db05 6/25/2012 3 
dev0db05 6/26/2012 6 
dev0db05 6/27/2012 9 
dev0db05 6/28/2012 12 

所以在这个例子中我想只有最高SumKB返回（12）dev0db05 我知道我必须做一个MAX，但不知道如何针对上述... 谢谢，可能过于简单，但我在一个损失...

对于下面的Linq查询，我该如何返回只有最高值，即259835919？ 谢谢！我是一个新手，并试图理解。

dim q = From c In Stuff _ 
where c.client= "dev0db05" _ 
where c.policy = "mcg-oracle-db" _ 
order by c.BackupDate _ 
group c by key = 0, c.Client,c.backupdate into g = group _ 
select ClientName = Client, BDate = DateTime.Parse(backupdate).ToShortDateString, SumKB = g.sum(Function(p) p.kilobytes) 

回报：

dev0db05 5/20/2012 163160396 
dev0db05 5/27/2012 235918153 
dev0db05 6/3/2012 259813647 
dev0db05 6/10/2012 259835151 
dev0db05 6/13/2012 23824838 
dev0db05 6/14/2012 259578160 
dev0db05 6/15/2012 23864427 
dev0db05 6/16/2012 259578127 
**dev0db05 6/17/2012 259835919** 
dev0db05 6/18/2012 23858283 
dev0db05 6/19/2012 259575055 
dev0db05 6/20/2012 33412 

Answer 1

好的下面是解决方案。这是最有效的方法吗？

Structure Stuff 
    public client as String 
    public BackupDate as DateTime 
    public Kilobytes as double 
    public sub new(client as string, BackupDate as DateTime, Kilobytes as double) 
    Me.client = client 
    Me.BackupDate = BackupDate 
    Me.Kilobytes = Kilobytes 
    end sub 
End Structure 

Sub Main 
Dim Stuffs as new list(Of Stuff) 
Stuffs.Add(new Stuff("dev0db05", "6/25/2012 7:00:35 AM",1)) 
Stuffs.Add(new Stuff("dev0db05", "6/25/2012 7:00:35 AM",1)) 
Stuffs.Add(new Stuff("dev0db05", "6/25/2012 7:00:35 AM",1)) 
Stuffs.Add(new Stuff("dev0db05", "6/26/2012 7:00:35 AM",2)) 
Stuffs.Add(new Stuff("dev0db05", "6/26/2012 7:00:35 AM",2)) 
Stuffs.Add(new Stuff("dev0db05", "6/26/2012 7:00:35 AM",2)) 
Stuffs.Add(new Stuff("dev0db06", "6/27/2012 7:00:35 AM",3)) 
Stuffs.Add(new Stuff("dev0db06", "6/27/2012 7:00:35 AM",3)) 
Stuffs.Add(new Stuff("dev0db06", "6/27/2012 7:00:35 AM",3)) 
Stuffs.Add(new Stuff("dev0db06", "6/28/2012 7:00:35 AM",4)) 
Stuffs.Add(new Stuff("dev0db06", "6/28/2012 7:00:35 AM",4)) 
Stuffs.Add(new Stuff("dev0db06", "6/28/2012 7:00:35 AM",4)) 
Stuffs.Add(new Stuff("dev0db09", "6/27/2012 7:00:35 AM",5)) 
Stuffs.Add(new Stuff("dev0db09", "6/27/2012 7:00:35 AM",5)) 
Stuffs.Add(new Stuff("dev0db09", "6/27/2012 7:00:35 AM",5)) 
Stuffs.Add(new Stuff("dev0db03", "6/28/2012 7:00:35 AM",6)) 
Stuffs.Add(new Stuff("dev0db03", "6/28/2012 7:00:35 AM",6)) 
Stuffs.Add(new Stuff("dev0db03", "6/28/2012 7:00:35 AM",6)) 

dim jj = From t1 In ((From t0 In Stuffs _ 
    Group t0 By t0.Client,t0.BackupDate Into g = Group _ 
    Select SumKB = CType(g.Sum(Function(p) p.Kilobytes),long?),Client,BDate = CStr(CDate(BackupDate)))) _ 
where t1.client = "dev0db05" _ 
Group t1 By t1.Client Into g = Group _ 
Select Client, MaxKB = CType(g.Max(Function(p) p.SumKB),long?) 

返回6 dev0db05