如何仅选择LINQ中具有最高日期的记录

Question

我有一个表'lasttraces',带有以下字段.

Id, AccountId, Version, DownloadNo, Date

数据如下所示:

28092|15240000|1.0.7.1782|2009040004731|2009-01-20 13:10:22.000
28094|61615000|1.0.7.1782|2009040007696|2009-01-20 13:11:38.000
28095|95317000|1.0.7.1782|2009040007695|2009-01-20 13:10:18.000
28101|15240000|1.0.7.1782|2009040004740|2009-01-20 14:10:22.000
28103|61615000|1.0.7.1782|2009040007690|2009-01-20 14:11:38.000
28104|95317000|1.0.7.1782|2009040007710|2009-01-20 14:10:18.000

在LINQ to SQL中,我怎样才能获得每个AccountId(具有最高日期的那个)的最后一个lasttrace？

Answer 1

如果您只想要每个帐户的最后日期,请使用以下内容:

var q = from n in table
        group n by n.AccountId into g
        select new {AccountId = g.Key, Date = g.Max(t=>t.Date)};

如果你想要整个记录:

var q = from n in table
        group n by n.AccountId into g
        select g.OrderByDescending(t=>t.Date).FirstOrDefault();

Answer 2

这是一个简单的方法

var lastPlayerControlCommand = this.ObjectContext.PlayerControlCommands
                                .Where(c => c.PlayerID == player.ID)
                                .OrderByDescending(t=>t.CreationTime)
                                .FirstOrDefault();

还看看这个很棒的LINQ地方 - LINQ to SQL Samples

Answer 3

如果你想要整个记录,这里是一个lambda方式:

var q = _context
             .lasttraces
             .GroupBy(s => s.AccountId)
             .Select(s => s.OrderByDescending(x => x.Date).FirstOrDefault());

Answer 4

可能是这样的：

var qry = from t in db.Lasttraces
          group t by t.AccountId into g
          orderby t.Date
          select new { g.AccountId, Date = g.Max(e => e.Date) };