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
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.
其他回答
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.
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;
}
下面是NuGet使用FastMember的2013年更新:
IEnumerable<SomeType> data = ...
DataTable table = new DataTable();
using(var reader = ObjectReader.Create(data)) {
table.Load(reader);
}
它使用FastMember的元编程API来获得最大的性能。如果你想将它限制到特定的成员(或强制执行顺序)，那么你也可以这样做:
IEnumerable<SomeType> data = ...
DataTable table = new DataTable();
using(var reader = ObjectReader.Create(data, "Id", "Name", "Description")) {
table.Load(reader);
}
FastMember是Marc Gravell的一个项目。它是金色的，完全是苍蝇!
是的，这个和这个完全相反;反射就足够了——或者如果你需要更快，2.0中的HyperDescriptor，或者3.5中的Expression。实际上，HyperDescriptor应该绰绰有余。
例如:
// remove "this" if not on C# 3.0 / .NET 3.5
public static DataTable ToDataTable<T>(this IList<T> data)
{
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, prop.PropertyType);
}
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);
}
table.Rows.Add(values);
}
return table;
}
现在，只需一行代码，你就可以让它比反射快很多很多倍(通过为对象类型T启用HyperDescriptor)。
编辑重新性能查询;下面是测试结果:
Vanilla 27179
Hyper 6997
我怀疑瓶颈已经从成员访问转移到数据表性能…我怀疑你在这方面还能提高多少……
代码:
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Diagnostics;
public class MyData
{
public int A { get; set; }
public string B { get; set; }
public DateTime C { get; set; }
public decimal D { get; set; }
public string E { get; set; }
public int F { get; set; }
}
static class Program
{
static void RunTest(List<MyData> data, string caption)
{
GC.Collect(GC.MaxGeneration, GCCollectionMode.Forced);
GC.WaitForPendingFinalizers();
GC.WaitForFullGCComplete();
Stopwatch watch = Stopwatch.StartNew();
for (int i = 0; i < 500; i++)
{
data.ToDataTable();
}
watch.Stop();
Console.WriteLine(caption + "\t" + watch.ElapsedMilliseconds);
}
static void Main()
{
List<MyData> foos = new List<MyData>();
for (int i = 0 ; i < 5000 ; i++ ){
foos.Add(new MyData
{ // just gibberish...
A = i,
B = i.ToString(),
C = DateTime.Now.AddSeconds(i),
D = i,
E = "hello",
F = i * 2
});
}
RunTest(foos, "Vanilla");
Hyper.ComponentModel.HyperTypeDescriptionProvider.Add(
typeof(MyData));
RunTest(foos, "Hyper");
Console.ReadLine(); // return to exit
}
}
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;
}
