从高效的数据结构来保存员工的活动?,我有一个 ActividadEmpleado 类型的列表,声明为:
public string Empleado { get; set; }
public DateTime Fecha { get; set; }
public string Actividad { get; set; }
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LINQ 查询变量以我需要的方式对结果重新排序,即按日期存储,然后按 ActividadEmpleado 和字符串存储。但是,var 类型不能传递给方法,所以搜索这个网站我发现我要么需要创建一个类来存储结果或修改 LINQ 变量以返回一个列表,但我在正确声明方面遇到了问题.
LINQ 变量是:
var queryActividades = listaActividad
.GroupBy(a => a.Fecha, (fecha, fechaActividades) => new
{
Fecha = fecha,
FechaActividades = fechaActividades
.GroupBy(a => a.Empleado, (nombreEmpleado, actividadesEmpleado) => new
{
Actividades = actividadesEmpleado,
NombreEmpleado = nombreEmpleado
})
.OrderBy(a => a.NombreEmpleado)
})
.OrderBy(a => a.Fecha);
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Visual Studio 说 queryActividades 是: …
可能重复:
LINQ to SQL:返回匿名类型?
这是我的代码:
class B
{
public int age { get; set; }
public string name { get; set; }
}
class Program
{
static List<whatshouldIwritehere> GetList()
{
List<B> list = new List<B>() {
new B(){ age = 10, name = "jaagu" },
new B(){ age = 20, name = "juggu" },
new B(){ age = 30, name = "jockey" },
new B(){ age = 40, name = "jaggu" },
};
return (from item in list
select new …Run Code Online (Sandbox Code Playgroud) 我正在使用枚举和LINQ分组来分组来自具有5列的DataTable的数据.现在我需要访问结果中的每个列数据.
编辑:
private IEnumerable<object> getItemsForDisplay()
{
var result = toDisplay.AsEnumerable()
.GroupBy(x => new
{
Col1 = x.Field<string>("rItem"),
Col2 = x.Field<string>("rMaterial"),
Col3 = x.Field<string>("rSpecs")
})
.Select(g => new
{
Col1 = g.Key.Col1,
Col2 = g.Key.Col2,
Col3 = g.Key.Col3,
Col4 = String.Join(", ", g.Select(row => row.Field<string>("rQuantity"))),
Col5 = String.Join(", ", g.Select(row => row.Field<string>("rOptional"))),
}).ToList();
return result;
}
//In another function
foreach (var item in result)
{
//item.tostring shows this: {"aaa","bbb","ccc","ddd","eee")
//turn it to array string or list to access "aaa".. etc etc …Run Code Online (Sandbox Code Playgroud) 接收此列表作为参数的方法的正确语法是什么?
var customList = await db.MyDbTable
.Select(x => new { x.Id, x.ParentId, x.Title })
.ToListAsync();
MyMethod(customList);
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这不起作用...
private void MyMethod(List<int, int, string> inputList)
{
// process the input list
return;
}
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