在下面的示例代码中，我收到了错误：
元素 TestSerializeDictionary123.Customer.CustomProperties vom Typ System.Collections.Generic.Dictionary`2 [System.String， mscorlib，版本= 2.0.0.0， 文化=中性， 公钥= b77a5c561934e089]，[System.Object的， mscorlib，版本= 2.0.0.0， 文化=中性， PublicKeyToken = b77a5c561934e089]]可以 不被序列化，因为它 实现IDictionary。
当我取出Dictionary属性时，它可以精细。
如何使用字典属性序列化此Customer对象？或者我可以使用哪种替换类型的词典可以序列化？
using System;
using System.Collections.Generic;
using System.Xml.Serialization;
using System.IO;
using System.Xml;
using System.Text;
namespace TestSerializeDictionary123
{
public class Program
{
static void Main(string[] args)
{
List<Customer> customers = Customer.GetCustomers();
Console.WriteLine("--- Serializing ------------------");
foreach (var customer in customers)
{
Console.WriteLine("Serializing " + customer.GetFullName() + "...");
string xml = XmlHelpers.SerializeObject<Customer>(customer);
Console.WriteLine(xml);
Console.WriteLine("Deserializing ...");
Customer customer2 = XmlHelpers.DeserializeObject<Customer>(xml);
Console.WriteLine(customer2.GetFullName());
Console.WriteLine("---");
}
Console.ReadLine();
}
}
public static class StringHelpers
{
public static String UTF8ByteArrayToString(Byte[] characters)
{
UTF8Encoding encoding = new UTF8Encoding();
String constructedString = encoding.GetString(characters);
return (constructedString);
}
public static Byte[] StringToUTF8ByteArray(String pXmlString)
{
UTF8Encoding encoding = new UTF8Encoding();
Byte[] byteArray = encoding.GetBytes(pXmlString);
return byteArray;
}
}
public static class XmlHelpers
{
public static string SerializeObject<T>(object o)
{
MemoryStream ms = new MemoryStream();
XmlSerializer xs = new XmlSerializer(typeof(T));
XmlTextWriter xtw = new XmlTextWriter(ms, Encoding.UTF8);
xs.Serialize(xtw, o);
ms = (MemoryStream)xtw.BaseStream;
return StringHelpers.UTF8ByteArrayToString(ms.ToArray());
}
public static T DeserializeObject<T>(string xml)
{
XmlSerializer xs = new XmlSerializer(typeof(T));
MemoryStream ms = new MemoryStream(StringHelpers.StringToUTF8ByteArray(xml));
XmlTextWriter xtw = new XmlTextWriter(ms, Encoding.UTF8);
return (T)xs.Deserialize(ms);
}
}
public class Customer
{
public int Id { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
public string Street { get; set; }
public string Location { get; set; }
public string ZipCode { get; set; }
public Dictionary<string,object> CustomProperties { get; set; }
private int internalValue = 23;
public static List<Customer> GetCustomers()
{
List<Customer> customers = new List<Customer>();
customers.Add(new Customer { Id = 1, FirstName = "Jim", LastName = "Jones", ZipCode = "23434" });
customers.Add(new Customer { Id = 2, FirstName = "Joe", LastName = "Adams", ZipCode = "12312" });
customers.Add(new Customer { Id = 3, FirstName = "Jack", LastName = "Johnson", ZipCode = "23111" });
customers.Add(new Customer { Id = 4, FirstName = "Angie", LastName = "Reckar", ZipCode = "54343" });
customers.Add(new Customer { Id = 5, FirstName = "Henry", LastName = "Anderson", ZipCode = "16623" });
return customers;
}
public string GetFullName()
{
return FirstName + " " + LastName + "(" + internalValue + ")";
}
}
}
答案 0 :(得分：14)
在我们的申请中，我们最终使用：
DataContractSerializer xs = new DataContractSerializer(typeof (T));
而不是：
XmlSerializer xs = new XmlSerializer(typeof (T));
解决了这个问题，因为DatacontractSerializer支持Dictionary。
另一种解决方案是XML Serializable Generic Dictionary解决方法也可以在上面的示例中使用，并且在使用它的人的链接上进行了长时间的讨论，可能对处理此问题的人有用。
答案 1 :(得分：9)
这是一个知道如何自行序列化的通用字典类：
public class XmlDictionary<T, V> : Dictionary<T, V>, IXmlSerializable {
[XmlType("Entry")]
public struct Entry {
public Entry(T key, V value) : this() { Key = key; Value = value; }
[XmlElement("Key")]
public T Key { get; set; }
[XmlElement("Value")]
public V Value { get; set; }
}
System.Xml.Schema.XmlSchema IXmlSerializable.GetSchema() {
return null;
}
void IXmlSerializable.ReadXml(System.Xml.XmlReader reader) {
this.Clear();
var serializer = new XmlSerializer(typeof(List<Entry>));
reader.Read(); // Why is this necessary?
var list = (List<Entry>)serializer.Deserialize(reader);
foreach (var entry in list) this.Add(entry.Key, entry.Value);
reader.ReadEndElement();
}
void IXmlSerializable.WriteXml(System.Xml.XmlWriter writer) {
var list = new List<Entry>(this.Count);
foreach (var entry in this) list.Add(new Entry(entry.Key, entry.Value));
XmlSerializer serializer = new XmlSerializer(list.GetType());
serializer.Serialize(writer, list);
}
}
答案 2 :(得分：4)
你不能（自己做不了，这太可怕了）; xml序列化程序不会知道如何处理object，因为它不包含有线格式的类型元数据。一个（hacky）选项是将这些all作为字符串流式传输以用于序列化，但是你需要编写许多额外的解析（等）代码。
答案 3 :(得分：1)
您可以改用Binary serialization。 （只需确保所有类都标记为[Serializable]。当然，它不是XML格式，但您没有将其列为要求：）
答案 4 :(得分：0)
我刚发现这个blog post by Rakesh Rajan描述了一种可能的解决方案：
通过使类型实现System.Xml.Serialization.IXmlSerializable类来覆盖XmlSerialization。在WriteXml方法中定义如何在XML中序列化对象，并定义如何从ReadXml方法中的xml字符串重新创建对象。
但是这不起作用，因为您的词典包含object而不是特定类型。
答案 5 :(得分：0)
如何将Customer类标记为DataContract，将其属性标记为DataMembers。 DataContract序列化器将为您进行序列化。
答案 6 :(得分：0)
尝试通过BinaryFormatter进行序列化
private void Deserialize()
{
try
{
var f_fileStream = File.OpenRead(@"dictionarySerialized.xml");
var f_binaryFormatter = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter();
myDictionary = (Dictionary<string, myClass>)f_binaryFormatter.Deserialize(f_fileStream);
f_fileStream.Close();
}
catch (Exception ex)
{
;
}
}
private void Serialize()
{
try
{
var f_fileStream = new FileStream(@"dictionarySerialized.xml", FileMode.Create, FileAccess.Write);
var f_binaryFormatter = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter();
f_binaryFormatter.Serialize(f_fileStream, myDictionary);
f_fileStream.Close();
}
catch (Exception ex)
{
;
}
}