序列化具有相同属性变体的对象
序列化具有相同属性变体的对象
提问于 2013-03-09 21:23:57
我有一个可以序列化对象的方法。
例如。
List<Titles> lstCrmTitleSer = new List<Titles>();我循环遍历字符串标题的集合，并将其添加到类型标题的列表中以序列化它们。
foreach (var ls in lstCrmTitles)
{
Titles t = new Titles();
t.title = ls;
lstCrmTitleSer.Add(t);
}这是我的对象和我用来序列化对象的方法。
static public void SerializeToXMLCollection(List<Titles> trainingTitles)
{
XmlSerializer serializer = new XmlSerializer(typeof(List<Titles>));
string path = string.Concat(@"C:\Users\Arian\Desktop\Titles.xml");
Console.WriteLine(path);
TextWriter textWriter = new StreamWriter(path);
serializer.Serialize(textWriter, trainingTitles);
textWriter.Close();
}
public string title { get; set; }问题是我的Xml文件显示如下
<Titles>
<title>title a</title>
</Titles>
<Titles>
<title>title b</title>
</Titles>
<Titles>
<title>title c</title>
</Titles> 如果我的xml按如下方式显示，我希望是什么
<Titles>
<title>title a</title>
<title>title b</title>
<title>title c</title>
</Titles>我如何调整我的代码来实现上述结果？
回答 1
Stack Overflow用户
回答已采纳
发布于 2013-03-09 21:37:55
第一。下面是一个可供您使用的通用扩展方法。我几年前写的，它可以序列化任何东西：
/// <summary>
/// <para>Serializes the specified System.Object and writes the XML document</para>
/// <para>to the specified file.</para>
/// </summary>
/// <typeparam name="T">This item's type</typeparam>
/// <param name="item">This item</param>
/// <param name="fileName">The file to which you want to write.</param>
/// <returns>true if successful, otherwise false.</returns>
public static bool XmlSerialize<T>(this T item, string fileName)
{
return item.XmlSerialize(fileName, true);
}
/// <summary>
/// <para>Serializes the specified System.Object and writes the XML document</para>
/// <para>to the specified file.</para>
/// </summary>
/// <typeparam name="T">This item's type</typeparam>
/// <param name="item">This item</param>
/// <param name="fileName">The file to which you want to write.</param>
/// <param name="removeNamespaces">
/// <para>Specify whether to remove xml namespaces.</para>para>
/// <para>If your object has any XmlInclude attributes, then set this to false</para>
/// </param>
/// <returns>true if successful, otherwise false.</returns>
public static bool XmlSerialize<T>(this T item, string fileName, bool removeNamespaces)
{
object locker = new object();
XmlSerializerNamespaces xmlns = new XmlSerializerNamespaces();
xmlns.Add(string.Empty, string.Empty);
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
settings.OmitXmlDeclaration = true;
lock (locker)
{
using (XmlWriter writer = XmlWriter.Create(fileName, settings))
{
if (removeNamespaces)
{
xmlSerializer.Serialize(writer, item, xmlns);
}
else { xmlSerializer.Serialize(writer, item); }
writer.Close();
}
}
return true;
}
/// <summary>
/// Serializes the specified System.Object and returns the serialized XML
/// </summary>
/// <typeparam name="T">This item's type</typeparam>
/// <param name="item">This item</param>
/// <param name="removeNamespaces">
/// <para>Specify whether to remove xml namespaces.</para>para>
/// <para>If your object has any XmlInclude attributes, then set this to false</para>
/// </param>
/// <returns>Serialized XML for specified System.Object</returns>
public static string XmlSerialize<T>(this T item, bool removeNamespaces = true)
{
object locker = new object();
XmlSerializerNamespaces xmlns = new XmlSerializerNamespaces();
xmlns.Add(string.Empty, string.Empty);
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
settings.OmitXmlDeclaration = true;
lock (locker)
{
StringBuilder stringBuilder = new StringBuilder();
using (StringWriter stringWriter = new StringWriter(stringBuilder))
{
using (XmlWriter xmlWriter = XmlWriter.Create(stringWriter, settings))
{
if (removeNamespaces)
{
xmlSerializer.Serialize(xmlWriter, item, xmlns);
}
else { xmlSerializer.Serialize(xmlWriter, item); }
return stringBuilder.ToString();
}
}
}
}第二。像这样使用它：
internal class Program
{
private static void Main(string[] args)
{
var list = new MyObject();
list.Titles.Add("title1");
list.Titles.Add("title2");
list.Titles.Add("title3");
string serialized = list.XmlSerialize();
//obviously here you would save to disk or whatever
Console.WriteLine(serialized);
Console.ReadLine();
}
}
[XmlRoot("Titles")]
public class MyObject
{
[XmlElement("title")]
public List<string> Titles { get; set; }
public MyObject()
{
Titles = new List<string>();
}
}第三..这段代码实在是太糟糕了。即使有了上面的建议，我仍然认为你需要做一些认真的重构。
祝好运
BONUS
将字符串反转换为您的类型的扩展：
/// <summary>
/// Deserializes the XML data contained by the specified System.String
/// </summary>
/// <typeparam name="T">The type of System.Object to be deserialized</typeparam>
/// <param name="s">The System.String containing XML data</param>
/// <returns>The System.Object being deserialized.</returns>
public static T XmlDeserialize<T>(this string s)
{
if (string.IsNullOrEmpty(s))
{
return default(T);
}
var locker = new object();
var stringReader = new StringReader(s);
var reader = new XmlTextReader(stringReader);
try
{
var xmlSerializer = new XmlSerializer(typeof(T));
lock (locker)
{
var item = (T)xmlSerializer.Deserialize(reader);
reader.Close();
return item;
}
}
catch
{
return default(T);
}
finally
{
reader.Close();
}
}像这样使用它：
var myObject = someSerializedString.XmlDeserialize<MyObject>()页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/15311177
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