2

この形式（Acc、sDate、Serial、Amount、...）のような6000000レコードの大きなテーブルがあります。Acc、date、serialはPKeyです。

私の問題を示すために、小さなコードを作成しました

public class Cheque 
{
    public string Account{ get; set; }
    public string Serial{ get; set; }   
    public string StartDate { get; set; }
    // ... public string Amount { get; set; }    ...
}

var list = new List<Cheque>();
list.Add(new Cheque() { Account= "1", Serial = "1", StartDate = "20080120"});
list.Add(new Cheque() { Account= "1", Serial= "2", StartDate = "20080120" });
list.Add(new Cheque() { Account= "1", Serial= "3", StartDate = "20080120" }); 
list.Add(new Cheque() { Account= "1", Serial= "4", StartDate = "20080120" }); 
// each acc have 100 to 300 record per date ,for simplicity 3 obj added

list.Add(new Cheque() { Account= "1", Serial= "1", StartDate = "20110120" });
list.Add(new Cheque() { Account= "1", Serial= "2", StartDate = "20110120" });

list.Add(new Cheque() { Account= "1", Serial= "1", StartDate = "20120120" });
list.Add(new Cheque() { Account= "1", Serial= "2", StartDate = "20120120" });
list.Add(new Cheque() { Account= "1", Serial= "3", StartDate = "20120120" });

list.Add(new Cheque() { Account= "2", Serial= "1", StartDate = "20100417" });
list.Add(new Cheque() { Account= "2", Serial= "2", StartDate = "20100417" });

list.Add(new Cheque() { Account= "2", Serial= "1", StartDate = "20120314" });

list.Add(new Cheque() { Account= "2", Serial= "1", StartDate = "20070301" });
list.Add(new Cheque() { Account= "2", Serial= "1", StartDate = "20070301" });
list.Add(new Cheque() { Account= "2", Serial= "1", StartDate = "20070301" });

期待されるリストは、各アカウントからの最も近い日付で設定された 2 つだけです

会計シリアル日

"1", "1", "20120120"   //first resultSet with Account= 1 
"1", "2", "20120120" 
"1", "3", "20120120"
"1", "1", "20110120"  //second resultset with Account=  1 
"1", "2", "20110120" 
"2", "1", "20120314"  //first resultSet with Account= 2 
"2", "1", "20100417" //second resultset with Account=  2 
"2", "2", "20100417"

plz は、linq を使用してこれをクエリする方法を教えてください。

4

4 に答える 4

2

トリックは、Account and Serialでグループ化することです。上位 2 つの日付を取得し、SelectMany によってリストを再び平坦化します。

list.GroupBy(x => new {x.Account, x.Serial})
.Select(g => new { FirstTwo = g
                   .GroupBy(x => x.StartDate).Select(x => x.FirstOrDefault())
                   .OrderByDescending(x => x.StartDate).Take(2)
                 })
.SelectMany(g => g.FirstTwo)
.OrderBy(x => x.Account)
    .ThenByDescending(x => x.StartDate)
        .ThenBy(x => x.Serial)

結果：

1   1   20120120
1   2   20120120
1   3   20120120
1   1   20110120
1   2   20110120
1   3   20110120
2   1   20120314
2   2   20120314
2   1   20100417
2   2   20100417
于 2013-09-15T13:08:08.500 に答える
0

最後に、望ましい結果を生み出すステートメントを 1 つ見つけました。

var result = (from cheque in list.OrderBy(a => a.Account).ThenByDescending(a => a.StartDate)
                            group cheque by new { cheque.Account, cheque.StartDate } into gr
                            //from ids in gr
                            group gr by gr.Key.Account into secondGrouping
                            from second in secondGrouping.Distinct().Take(2)
                                  from Cheque f in second
                                  select f 
                            ).ToList<Cheque>();
于 2013-09-18T12:20:41.457 に答える