为每个ID选择具有最大日期的值

这是表“历史”

id value date 1 1 01/01/2017 20:20:20 1 2 02/01/2017 20:20:20 1 3 03/01/2017 20:20:20 2 5 01/01/2017 20:20:20 2 6 02/01/2017 20:20:20 

如何使用linq为每个id选择最大值

 context.History .GroupBy(x => x.id) ?? .SelectOnlyWithMax(z => z.date) ?? 

结果只有两个对象

 id value date 1 3 03/01/2017 20:20:20 2 6 02/01/2017 20:20:20 

Answer 1

如果您希望每个Id日期最长的整行，您可以使用以下代码（使用LinqPad编写）。 如果你只想要Id ，你可以使用@BurnsBA的答案，因为它会更有效率。

 void Main() { var data = new List { new Record(){Id=1, Value=1, Date=new DateTime(2017,1,1)}, new Record(){Id=1, Value=2, Date=new DateTime(2017,2,1)}, new Record(){Id=1, Value=3, Date=new DateTime(2017,3,1)}, new Record(){Id=2, Value=5, Date=new DateTime(2017,1,1)}, new Record(){Id=2, Value=6, Date=new DateTime(2017,2,1)}, }; var query = data.GroupBy(d => d.Id) .SelectMany(g => g.OrderByDescending(d => d.Date) .Take(1)); query.Dump(); } public class Record { public int Id { get; set; } public int Value { get; set; } public DateTime Date { get; set; } } 

结果：

首先按Id分组，然后按Date按降序对组内的项目进行排序，然后返回第一个， SelectMany然后展平列表。

Answer 2

我建议使用MoreLINQ的MaxBy函数，即：

 context.History.GroupBy( x => x.id ).Select( x => x.MaxBy( y => y.date) ) 

Answer 3

 public class History { public int id { get; set; } public int value { get; set; } public DateTime date { get; set; } } // setup: var values = new List(); values.Add(new History() { id = 1, value = 1, date = DateTime.Parse("01/01/2017 20:20:20") }); values.Add(new History() { id = 1, value = 2, date = DateTime.Parse("02/01/2017 20:20:20") }); values.Add(new History() { id = 1, value = 3, date = DateTime.Parse("03/01/2017 20:20:20") }); values.Add(new History() { id = 2, value = 5, date = DateTime.Parse("01/01/2017 20:20:20") }); values.Add(new History() { id = 2, value = 6, date = DateTime.Parse("02/01/2017 20:20:20") }); // result : values.GroupBy( x => x.id, y => y.date, // Below, dates will be enumerable (id, dates) => new { id = id, date = dates.Max() } ) // returns enumerable collection of anonymous type: { { id = 1, date = [3/1/2017 8:20:20 PM] }, { id = 2, date = [2/1/2017 8:20:20 PM] } }