Question

我从MySQL表中得到结果如下。每次数据的结果都不同，我需要将它分组以分离新的List＆lt; int＆gt;。 DataTable中的数据列表：

id_task id_supplier id_source_ps
1   77  3
2   71  3
3   77  1
4   77  2
5   72  3
6   72  2
7   77  3
8   72  3
9   71  1
10  72  3

将id_task分组到新列表的条件与id_supplier和id_source_ps相同！

什么方法（LINQ？）用于得到结果 - 如本例所示 - 七个新List＆lt; int＆gt;

name of list    id_task_in
group1  9
group2  2
group3  6
group4  5,8,10
group5  3
group6  4
group7  1,7

Answer 1

您正在寻找GroupBy。您需要GroupBy您的密钥，然后Select您的结果。以下是使用您的数据的示例 -

public class Program
{
    public static void Main(string[] args)
    {
        var list = new List<Item>
        {
            new Item { task = 1, supplier = 77, source = 3 },
            new Item { task = 2, supplier = 71, source = 3 },
            new Item { task = 3, supplier = 77, source = 1 },
            new Item { task = 4, supplier = 77, source = 2 },
            new Item { task = 5, supplier = 72, source = 3 },
            new Item { task = 6, supplier = 72, source = 2 },
            new Item { task = 7, supplier = 77, source = 3 },
            new Item { task = 8, supplier = 72, source = 3 },
            new Item { task = 9, supplier = 71, source = 1 },
            new Item { task = 10, supplier = 72, source = 3 }
        };

        var groupBy = 
            list
            .GroupBy(x => new { first = x.source, second = x.supplier})
            .Select(x => new { name = x.Key, items = x.Count() });

        foreach (var g in groupBy)
        {
            Console.WriteLine(g);
        }        
    }        
}

public class Item
{
    public int task;
    public int supplier;
    public int source;
}

打印

{ name = { first = 3, second = 77 }, items = 2 }
{ name = { first = 3, second = 71 }, items = 1 }
{ name = { first = 1, second = 77 }, items = 1 }
{ name = { first = 2, second = 77 }, items = 1 }
{ name = { first = 3, second = 72 }, items = 3 }
{ name = { first = 2, second = 72 }, items = 1 }
{ name = { first = 1, second = 71 }, items = 1 }

您可以将此示例用作起点。要完成您的需要，只需将.Count()更改为.Aggregate()，即可返回任务ID的列表，而不是任务ID的数量。如果您愿意，也可以更改密钥以匹配“组1”“组2”，但您需要为每个组编写一些标准方法（例如"group"+(index+1).ToString())。

Answer 2

以汤姆的回答为基础：

public static void Main(string[] args)
{
    var list = new List<Item>
    {
        new Item { task = 1, supplier = 77, source = 3 },
        new Item { task = 2, supplier = 71, source = 3 },
        new Item { task = 3, supplier = 77, source = 1 },
        new Item { task = 4, supplier = 77, source = 2 },
        new Item { task = 5, supplier = 72, source = 3 },
        new Item { task = 6, supplier = 72, source = 2 },
        new Item { task = 7, supplier = 77, source = 3 },
        new Item { task = 8, supplier = 72, source = 3 },
        new Item { task = 9, supplier = 71, source = 1 },
        new Item { task = 10, supplier = 72, source = 3 }
    };

    var groupBy = list.GroupBy(x => new { first = x.supplier, second = x.source})
      .OrderBy(x=>x.Key.first).ThenBy(x=>x.Key.second)
      .Select((x,i) => new { name = "group"+(i+1), items = x.Select(y=>y.task) });

    foreach (var g in groupBy)
    {
        Console.WriteLine("{0} {1}",g.name,String.Join(",",g.items.Select(x=>x.ToString())));
    }        
}        

public class Item
{
    public int task;
    public int supplier;
    public int source;
}

将返回：

group1 9
group2 2
group3 6
group4 5,8,10
group5 3
group6 4
group7 1,7

我将最终输出保留为IEnumerable<string,IEnumerable<int>>，因为我通常更喜欢这样。您可以轻松地将string.join移动到select中，如果您希望将其改为IEnumerable<string,string>。

分割结果来自SQL以分隔List <int>

2 个答案: