我的应用程序中有以下代码:
[Serializable]
public class Class
{
private string name = "";
private List<Property> property = new List<Property>();
[XmlAttribute]
public string Name
{
get { return name; }
set { name = value; }
}
[XmlArray(ElementName = "Properties")]
public List<Property> Property
{
get { return property; }
set { property = value; }
}
public Class() { }
}
[Serializable]
public class Property
{
private string name = "";
private object value;
[XmlAttribute]
public string Name
{
get { return name; }
set { name = value; }
}
[XmlElement]
public object Value
{
get { return this.value; }
set { this.value = value; }
}
public Property() { }
}
class Program
{
static void Main(string[] args)
{
List<Class> classList = GetClassList();
SerializeConfig("test.xml", classList);
}
private static List<Class> GetClassList()
{
return new List<Class>(){
new Class()
{
Name = "MyFirstClass",
Property = new List<Property>() {
new Property(){ Name = "StringProperty", Value = "Simple1"},
new Property(){ Name = "IntProperty", Value = 3},
new Property(){ Name = "DoubleProperty", Value = 5.5},
new Property(){ Name = "ListProperty", Value = new List<int>(){
1,2,3,4,5
}
},
}
}
};
}
public static List<Class> DeserializeConfig(string configPath)
{
if (!File.Exists(configPath)) return null;
try
{
XmlSerializer serializer = new XmlSerializer(typeof(List<Class>), new XmlRootAttribute("Classes"));
using (FileStream fs = File.Open(configPath, FileMode.Open))
{
return (List<Class>)serializer.Deserialize(fs);
}
}
catch (Exception)
{
return null;
}
}
public static bool SerializeConfig(string path, List<Class> config)
{
try
{
XmlSerializer serializer = new XmlSerializer(typeof(List<Class>), new XmlRootAttribute("Classes"));
using (FileStream fs = File.Open(path, FileMode.Create))
{
serializer.Serialize(fs, config);
}
}
catch (Exception)
{
return false;
}
return true;
}
}
如果您运行此代码,则会收到System.InvalidOperationException
,其中包含以下消息:
{System.InvalidOperationException: There was an error generating the XML document. ---> System.InvalidOperationException: The type System.Collections.Generic.List`1[[System.Int32, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089]] may not be used in this context.
at System.Xml.Serialization.XmlSerializationWriter.WriteTypedPrimitive(String name, String ns, Object o, Boolean xsiType)
at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationWriterList1.Write1_Object(String n, String ns, Object o, Boolean isNullable, Boolean needType)
at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationWriterList1.Write2_Property(String n, String ns, Property o, Boolean isNullable, Boolean needType)
at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationWriterList1.Write3_Class(String n, String ns, Class o, Boolean isNullable, Boolean needType)
at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationWriterList1.Write4_Classes(Object o)
--- End of inner exception stack trace ---
at System.Xml.Serialization.XmlSerializer.Serialize(XmlWriter xmlWriter, Object o, XmlSerializerNamespaces namespaces, String encodingStyle, String id)
at System.Xml.Serialization.XmlSerializer.Serialize(Stream stream, Object o, XmlSerializerNamespaces namespaces)
at System.Xml.Serialization.XmlSerializer.Serialize(Stream stream, Object o)
at SerializeTest.Program.SerializeConfig(String path, List`1 config) in [...]\Program.cs:line 102}
为什么我不能以这种方式序列化List?我如何建立一个变通方法?
答案 0 :(得分:1)
好的,我找到了解决方案。我必须添加自定义类型。
XmlSerializer serializer = new XmlSerializer(typeof(List<Class>), null, new Type[] { typeof(List<int>) }, new XmlRootAttribute("Classes"),"");
所以我必须从我的列表中收集所有类型:
private static Type[] CollectTypesFromCollection(List<Class> collection)
{
List<Type> types = new List<Type>();
foreach (Class item in collection)
foreach (Property property in item.Property)
if (!types.Contains(property.Value.GetType()))
types.Add(property.Value.GetType());
return types.ToArray();
}
可以像这样使用序列化器:
XmlSerializer serializer = new XmlSerializer(typeof(List<Class>),null,CollectTypesFromCollection(config), new XmlRootAttribute("Classes"),"");
答案 1 :(得分:-1)
由于XML Serialization
创建的方式,我建议您在进入发生错误的List
之前先查看一下。