为每个ID选择具有最大日期的值 [英] Select values with max date for each ID

问题描述

这是表格历史"

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

How with linq select max values for each id

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

推荐答案

如果希望每个Id的整个行都具有最高的日期，则可以使用以下代码(用LinqPad编写).如果只需要Id，则可以使用@BurnsBA的答案，因为这样会更有效.

If you want the entire row with the highest date for each Id, you can use the following code (written with LinqPad). If you just want the Id, you can use @BurnsBA's answer, as it will be slightly more efficient.

void Main()
{
    var data = new List<Record>
    {
        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:

首先将其按Id分组，然后按Date降序对组中的项目进行排序，然后返回第一个，SelectMany然后对列表进行展平.

First it groups by Id, then sorts the items within the group by Date in descending order, and returns the first one, SelectMany then flattens the list.

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