Question

对象object转换为XML

如何将对象转换为XML文件

若以下回答无法解决问题，邀请你更新回答

Answer 1

/// <summary>
/// SimpleXmlConverter 将对象object转换为XML。支持N层对象嵌入，并且支持IList<>接口。
/// </summary>
public static class SimpleXmlConverter
{
public static string XmlSerializeObject(object target)
{
string typeName = TypeHelper.GetClassSimpleName(target.GetType());
XmlDocument xmlDoc = new XmlDocument();
XmlElement objNode = xmlDoc.CreateElement(typeName); foreach (PropertyInfo pro in target.GetType().GetProperties())
{
if ((!pro.CanRead))
{
continue;
} SimpleXmlConverter.DoXmlSerializeObject(objNode, pro, pro.GetValue(target, null));
} return objNode.OuterXml;
} private static void DoXmlSerializeObject(XmlNode curNode, PropertyInfo pro, object val)
{
bool isGenericList = pro.PropertyType.IsGenericType && (pro.PropertyType.GetGenericTypeDefinition() == typeof(IList<>));
if (isGenericList)
{
string listName = pro.Name;
XmlNode listNode = curNode.OwnerDocument.CreateElement(listName);
curNode.AppendChild(listNode); Type listElementType = pro.PropertyType.GetGenericArguments()[0];
if (TypeHelper.IsSimpleType(listElementType))
{
foreach (object element in (IEnumerable)val)
{
XmlHelper.SetPropertyValue(listNode, "value", element.ToString(), XmlPropertyPosition.ChildNode, false);
}
}
else
{
foreach (object element in (IEnumerable)val)
{
string newObjName = TypeHelper.GetClassSimpleName(listElementType);
XmlNode newObjNode = curNode.OwnerDocument.CreateElement(newObjName);
listNode.AppendChild(newObjNode); PropertyInfo[] childPros = listElementType.GetProperties();
foreach (PropertyInfo childPro in childPros)
{
if ((!childPro.CanRead))
{
continue;
} object childVal = childPro.GetValue(element, null);
SimpleXmlConverter.DoXmlSerializeObject(newObjNode, childPro, childVal);
}
}
}
}
else if (TypeHelper.IsSimpleType(pro.PropertyType))
{
string valStr = (val == null ? "" : val.ToString());
XmlHelper.SetPropertyValue(curNode, pro.Name, valStr);
}
else
{
string newObjName = TypeHelper.GetClassSimpleName(pro.PropertyType);
XmlNode newObjNode = curNode.OwnerDocument.CreateElement(newObjName);
curNode.AppendChild(newObjNode); PropertyInfo[] childPros = pro.PropertyType.GetProperties();
foreach (PropertyInfo childPro in childPros)
{
if ((!childPro.CanRead))
{
continue;
} SimpleXmlConverter.DoXmlSerializeObject(newObjNode, childPro, childPro.GetValue(val, null));
}
} }
}

Answer 2

根据你的object类型来决定怎么转换。问题不甚明确。