我有几个方法返回不同的泛型列表。

在。net中存在任何类静态方法或将任何列表转换为数据表?我唯一能想到的是使用反射来做到这一点。

如果我有这个:

List<Whatever> whatever = new List<Whatever>();

(这下代码当然不工作,但我想有这样的可能性:

DataTable dt = (DataTable) whatever;

当前回答

 Dim counties As New List(Of County)
 Dim dtCounties As DataTable
 dtCounties = _combinedRefRepository.Get_Counties()
 If dtCounties.Rows.Count <> 0 Then
    For Each row As DataRow In dtCounties.Rows
      Dim county As New County
      county.CountyId = row.Item(0).ToString()
      county.CountyName = row.Item(1).ToString().ToUpper()
      counties.Add(county)
    Next
    dtCounties.Dispose()
 End If

其他回答

如果你使用VB。NET,那么这个类就会完成这项工作。

Imports System.Reflection
''' <summary>
''' Convert any List(Of T) to a DataTable with correct column types and converts Nullable Type values to DBNull
''' </summary>

Public Class ConvertListToDataset

    Public Function ListToDataset(Of T)(ByVal list As IList(Of T)) As DataTable

        Dim dt As New DataTable()
        '/* Create the DataTable columns */
        For Each pi As PropertyInfo In GetType(T).GetProperties()
            If pi.PropertyType.IsValueType Then
                Debug.Print(pi.Name)
            End If
            If IsNothing(Nullable.GetUnderlyingType(pi.PropertyType)) Then
                dt.Columns.Add(pi.Name, pi.PropertyType)
            Else
                dt.Columns.Add(pi.Name, Nullable.GetUnderlyingType(pi.PropertyType))
            End If
        Next

        '/* Populate the DataTable with the values in the Items in List */
        For Each item As T In list
            Dim dr As DataRow = dt.NewRow()
            For Each pi As PropertyInfo In GetType(T).GetProperties()
                dr(pi.Name) = IIf(IsNothing(pi.GetValue(item)), DBNull.Value, pi.GetValue(item))
            Next
            dt.Rows.Add(dr)
        Next
        Return dt

    End Function

End Class
 Dim counties As New List(Of County)
 Dim dtCounties As DataTable
 dtCounties = _combinedRefRepository.Get_Counties()
 If dtCounties.Rows.Count <> 0 Then
    For Each row As DataRow In dtCounties.Rows
      Dim county As New County
      county.CountyId = row.Item(0).ToString()
      county.CountyName = row.Item(1).ToString().ToUpper()
      counties.Add(county)
    Next
    dtCounties.Dispose()
 End If

2019年的答案,如果你正在使用。net Core——使用Nuget ToDataTable库。优点:

性能优于FastMember 还可以创建结构化的SqlParameters作为SQL Server表值参数

免责声明-我是ToDataTable的作者

性能——我扩展了一些Benchmark . net测试,并将它们包含在ToDataTable repo中。结果如下:

创建100,000行数据表:

                           MacOS         Windows
Reflection                 818.5 ms      818.3 ms
FastMember from           1105.5 ms      976.4 ms
 Mark's answer
Improved FastMember        524.6 ms      456.4 ms
ToDataTable                449.0 ms      376.5 ms

Marc回答中建议的FastMember方法的性能似乎比Mary使用反射的回答差,但我使用FastMember TypeAccessor滚动了另一个方法,它的性能要好得多。尽管如此,ToDataTable包的性能还是优于其他包。

我意识到这里已经关门有一段时间了;然而,我有一个解决这个特定问题的解决方案,但需要稍微扭转:列和数据表需要预先定义/已经实例化。然后我只需将类型插入到数据表中。

下面是我所做的一个例子:

public static class Test
{
    public static void Main()
    {
        var dataTable = new System.Data.DataTable(Guid.NewGuid().ToString());

        var columnCode = new DataColumn("Code");
        var columnLength = new DataColumn("Length");
        var columnProduct = new DataColumn("Product");

        dataTable.Columns.AddRange(new DataColumn[]
            {
                columnCode,
                columnLength,
                columnProduct
            });

        var item = new List<SomeClass>();

        item.Select(data => new
        {
            data.Id,
            data.Name,
            data.SomeValue
        }).AddToDataTable(dataTable);
    }
}

static class Extensions
{
    public static void AddToDataTable<T>(this IEnumerable<T> enumerable, System.Data.DataTable table)
    {
        if (enumerable.FirstOrDefault() == null)
        {
            table.Rows.Add(new[] {string.Empty});
            return;
        }

        var properties = enumerable.FirstOrDefault().GetType().GetProperties();

        foreach (var item in enumerable)
        {
            var row = table.NewRow();
            foreach (var property in properties)
            {
                row[property.Name] = item.GetType().InvokeMember(property.Name, BindingFlags.GetProperty, null, item, null);
            }
            table.Rows.Add(row);
        }
    }
}
List<YourModel> data = new List<YourModel>();
DataTable dataTable = Newtonsoft.Json.JsonConvert.DeserializeObject<DataTable>(Newtonsoft.Json.JsonConvert.SerializeObject(data));