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

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);
        }
    }
}

其他回答

我不得不修改Marc Gravell的示例代码来处理可空类型和空值。我在下面附上了一个工作版本。谢谢你马克。

public static DataTable ToDataTable<T>(this IList<T> data)
{
    PropertyDescriptorCollection properties = 
        TypeDescriptor.GetProperties(typeof(T));
    DataTable table = new DataTable();
    foreach (PropertyDescriptor prop in properties)
        table.Columns.Add(prop.Name, Nullable.GetUnderlyingType(prop.PropertyType) ?? prop.PropertyType);
    foreach (T item in data)
    {
        DataRow row = table.NewRow();
        foreach (PropertyDescriptor prop in properties)
             row[prop.Name] = prop.GetValue(item) ?? DBNull.Value;
        table.Rows.Add(row);
    }
    return table;
}

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)));
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.

另一种方法是:

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

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);
        }
    }
}