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

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;
}

其他回答

public DataTable ConvertToDataTable<T>(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;
}

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

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();

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;
}
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.