我有一个object {System.Collections.Generic.List<object>}
,里面包含1000 object {DynamicData}
,每个{4}包含4个键和值,还有一个List
,其中包含2个键和值。
我需要将此对象序列化为XML文件,我尝试了正常的序列化,但它给了我这个异常= The type DynamicData was not expected
,我如何序列化这个对象?
以下是代码:
//output is the name of my object
XmlSerializer xsSubmit = new XmlSerializer(output.GetType());
var xml = "";
using (var sww = new StringWriter())
{
using (XmlWriter writers = XmlWriter.Create(sww))
{
try
{
xsSubmit.Serialize(writers, output);
}
catch (Exception ex)
{
throw;
}
xml = sww.ToString(); // Your XML
}
}
我可以创建逐行逐行写入的xml文件,但我想要更快更快的代码。 我的对象的结构是这样的:
output (count 1000)
[0]
Costumer - "Costumername"
DT - "Date"
Key - "Key"
Payment - "x"
[0]
Adress - "x"
Number - "1"
[1]...
[2]...
答案 0 :(得分:2)
您可以使用IXmlSerializable
[Serializable]
public class ObjectSerialize : IXmlSerializable
{
public List<object> ObjectList { get; set; }
public XmlSchema GetSchema()
{
return new XmlSchema();
}
public void ReadXml(XmlReader reader)
{
}
public void WriteXml(XmlWriter writer)
{
foreach (var obj in ObjectList)
{
//Provide elements for object item
writer.WriteStartElement("Object");
var properties = obj.GetType().GetProperties();
foreach (var propertyInfo in properties)
{
//Provide elements for per property
writer.WriteElementString(propertyInfo.Name, propertyInfo.GetValue(obj).ToString());
}
writer.WriteEndElement();
}
}
}
<强>用法强>
var output = new List<object>
{
new { Sample = "Sample" }
};
var objectSerialize = new ObjectSerialize
{
ObjectList = output
};
XmlSerializer xsSubmit = new XmlSerializer(typeof(ObjectSerialize));
var xml = "";
using (var sww = new StringWriter())
{
using (XmlWriter writers = XmlWriter.Create(sww))
{
try
{
xsSubmit.Serialize(writers, objectSerialize);
}
catch (Exception ex)
{
throw;
}
xml = sww.ToString(); // Your XML
}
}
<强>输出强>
<?xml version="1.0" encoding="utf-16"?>
<ObjectSerialize>
<Object>
<Sample>Sample</Sample>
</Object>
</ObjectSerialize>
注意:如果要使用相同类型反序列化,请注意 (ObjectSerialize)你应该提供
ReadXml
。如果你愿意的话 指定架构,您也应该提供GetSchema
。