如果你的类中有属性，这行代码是OK的!!

PropertyDescriptorCollection props =
            TypeDescriptor.GetProperties(typeof(T));

但如果你有所有的公共字段，那么使用这个:

public static DataTable ToDataTable<T>(  IList<T> data)
        {
        FieldInfo[] myFieldInfo;
        Type myType = typeof(T);
        // Get the type and fields of FieldInfoClass.
        myFieldInfo = myType.GetFields(BindingFlags.NonPublic | BindingFlags.Instance
            | BindingFlags.Public);

        DataTable dt = new DataTable();
        for (int i = 0; i < myFieldInfo.Length; i++)
            {
            FieldInfo property = myFieldInfo[i];
            dt.Columns.Add(property.Name, property.FieldType);
            }
        object[] values = new object[myFieldInfo.Length];
        foreach (T item in data)
            {
            for (int i = 0; i < values.Length; i++)
                {
                values[i] = myFieldInfo[i].GetValue(item);
                }
            dt.Rows.Add(values);
            }
        return dt;
        }

原来的答案是从上面，我只是编辑使用字段而不是属性

然后这样使用它

 DataTable dt = new DataTable();
            dt = ToDataTable(myBriefs);
            gridData.DataSource = dt;
            gridData.DataBind();

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。

List<YourModel> data = new List<YourModel>();
DataTable dataTable = Newtonsoft.Json.JsonConvert.DeserializeObject<DataTable>(Newtonsoft.Json.JsonConvert.SerializeObject(data));

要将泛型列表转换为数据表，可以使用DataTableGenerator

这个库允许您将列表转换为具有多种功能的数据表，如

翻译数据表头 指定要显示的列

另一种方法是:

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

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.