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<object> Basket;

string json = JsonConvert.SerializeObject(Basket, Formatting.Indented);
DataTable dtUsingMethod = Business.GetJSONToDataTableUsingNewtonSoftDll(json);



public static DataTable GetJSONToDataTableUsingNewtonSoftDll(string JSONData)
{
    DataTable dt = (DataTable)JsonConvert.DeserializeObject(JSONData, (typeof(DataTable)));
    return dt;
}

  private DataTable CreateDataTable(IList<T> item)
        {
            Type type = typeof(T);
            var properties = type.GetProperties();

            DataTable dataTable = new DataTable();
            foreach (PropertyInfo info in properties)
            {
                dataTable.Columns.Add(new DataColumn(info.Name, Nullable.GetUnderlyingType(info.PropertyType) ?? info.PropertyType));
            }

            foreach (T entity in item)
            {
                object[] values = new object[properties.Length];
                for (int i = 0; i < properties.Length; i++)
                {
                    values[i] = properties[i].GetValue(entity);
                }

                dataTable.Rows.Add(values);
            }
            return dataTable;
        }

另一种方法是:

  List<WhateEver> lst = getdata();
  string json = Newtonsoft.Json.JsonConvert.SerializeObject(lst);
  DataTable pDt = JsonConvert.DeserializeObject<DataTable>(json);

我还必须想出一个替代解决方案，因为这里列出的选项都不适合我。我使用了一个IEnumerable，返回一个IEnumerable，属性不能被枚举。这招奏效了:

// remove "this" if not on C# 3.0 / .NET 3.5
public static DataTable ConvertToDataTable<T>(this IEnumerable<T> data)
{
    List<IDataRecord> list = data.Cast<IDataRecord>().ToList();

    PropertyDescriptorCollection props = null;
    DataTable table = new DataTable();
    if (list != null && list.Count > 0)
    {
        props = TypeDescriptor.GetProperties(list[0]);
        for (int i = 0; i < props.Count; i++)
        {
            PropertyDescriptor prop = props[i];
            table.Columns.Add(prop.Name, Nullable.GetUnderlyingType(prop.PropertyType) ?? prop.PropertyType);
        }
    }
    if (props != null)
    {
        object[] values = new object[props.Count];
        foreach (T item in data)
        {
            for (int i = 0; i < values.Length; i++)
            {
                values[i] = props[i].GetValue(item) ?? DBNull.Value;
            }
            table.Rows.Add(values);
        }
    }
    return table;
}

这是清单上的另一个。Cinchoo ETL -一个将枚举转换为数据表的开源库。

List<Whatever> whatever = new List<Whatever>();
var dt = whatever.AsDataTable();

免责声明:我是这个库的作者。