我认为它更方便和容易使用。

   List<Whatever> _lobj= new List<Whatever>(); 
    var json = JsonConvert.SerializeObject(_lobj);
                DataTable dt = (DataTable)JsonConvert.DeserializeObject(json, (typeof(DataTable)));

这是一个简单的混合溶液。 它适用于Nullable类型。

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

我自己编写了一个小型库来完成这个任务。它只在将对象类型转换为数据表时第一次使用反射。它发出一个方法，该方法将完成转换对象类型的所有工作。

它燃烧得很快。你可以在这里找到它:

ToDataTable()在MoreLinq上的GitHub 原:ModelShredder在GoogleCode，被合并到MoreLinq，从GoogleCode移动到GitHub

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。

您可以尝试以下内容

public static DataTable GetDataTableFromObjects(object[] objects)
{
    if (objects != null && objects.Length > 0)
    {
        Type t = objects[0].GetType();
        DataTable dt = new DataTable(t.Name);
        foreach (PropertyInfo pi in t.GetProperties())
        {
            dt.Columns.Add(new DataColumn(pi.Name));
        }
        foreach (var o in objects)
        {
            DataRow dr = dt.NewRow();
            foreach (DataColumn dc in dt.Columns)
            {
                dr[dc.ColumnName] = o.GetType().GetProperty(dc.ColumnName).GetValue(o, null);
            }
            dt.Rows.Add(dr);
        }
        return dt;
    }
    return null;
}

It's also possible through XmlSerialization.
The idea is - serialize to `XML` and then `readXml` method of `DataSet`.

I use this code (from an answer in SO, forgot where)

        public static string SerializeXml<T>(T value) where T : class
    {
        if (value == null)
        {
            return null;
        }

        XmlSerializer serializer = new XmlSerializer(typeof(T));

        XmlWriterSettings settings = new XmlWriterSettings();

        settings.Encoding = new UnicodeEncoding(false, false);
        settings.Indent = false;
        settings.OmitXmlDeclaration = false;
        // no BOM in a .NET string

        using (StringWriter textWriter = new StringWriter())
        {
            using (XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings))
            {
               serializer.Serialize(xmlWriter, value);
            }
            return textWriter.ToString();
        }
    }

so then it's as simple as:

            string xmlString = Utility.SerializeXml(trans.InnerList);

        DataSet ds = new DataSet("New_DataSet");
        using (XmlReader reader = XmlReader.Create(new StringReader(xmlString)))
        { 
            ds.Locale = System.Threading.Thread.CurrentThread.CurrentCulture;
            ds.ReadXml(reader); 
        }

Not sure how it stands against all the other answers to this post, but it's also a possibility.