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;
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
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();
其他回答
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)));
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();
Marc Gravell的回答,但是用VB。网
Public Shared Function ToDataTable(Of T)(data As IList(Of T)) As DataTable
Dim props As PropertyDescriptorCollection = TypeDescriptor.GetProperties(GetType(T))
Dim table As New DataTable()
For i As Integer = 0 To props.Count - 1
Dim prop As PropertyDescriptor = props(i)
table.Columns.Add(prop.Name, prop.PropertyType)
Next
Dim values As Object() = New Object(props.Count - 1) {}
For Each item As T In data
For i As Integer = 0 To values.Length - 1
values(i) = props(i).GetValue(item)
Next
table.Rows.Add(values)
Next
Return table
End Function
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.
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;
}