使用bog标准的System.Xml.Serialization.XmlSerializer,我正在序列化一个类继承自另一个的对象。检查生成的XML,根节点被赋予属性“p1:type”和“xmlns:p1”:
<ApiSubmission ApiVersion="1" CustId="100104" p1:type="OrderConfirmationApiSubmission"
xmlns:p1="http://www.w3.org/2001/XMLSchema-instance">
...
</ApiSubmission>
有没有一种很好的方法来删除这些属性?
答案 0 :(得分:0)
因此,在最初提出这个问题的5年后,我遇到了同样的问题,对没有人回答感到失望。搜索后,我将一些东西拼凑在一起,使我可以在派生类中删除type属性。
internal static string SerializeObject(object objectToSerialize, bool OmitXmlDeclaration = true, System.Type type = null, bool OmitType = false, bool RemoveAllNamespaces = true)
{
XmlSerializer x;
string output;
if (type != null)
{
x = new XmlSerializer(type);
}
else
{
x = new XmlSerializer(objectToSerialize.GetType());
}
XmlWriterSettings settings = new XmlWriterSettings() { Indent = false, OmitXmlDeclaration = OmitXmlDeclaration, NamespaceHandling = NamespaceHandling.OmitDuplicates };
using (StringWriter swriter = new StringWriter())
using (XmlWriter xmlwriter = XmlWriter.Create(swriter, settings))
{
x.Serialize(xmlwriter, objectToSerialize);
output = swriter.ToString();
}
if (RemoveAllNamespaces || OmitType)
{
XDocument doc = XDocument.Parse(output);
if (RemoveAllNamespaces)
{
foreach (var element in doc.Root.DescendantsAndSelf())
{
element.Name = element.Name.LocalName;
element.ReplaceAttributes(GetAttributesWithoutNamespace(element));
}
}
if (OmitType)
{
foreach (var node in doc.Descendants().Where(e => e.Attribute("type") != null))
{
node.Attribute("type").Remove();
}
}
output = doc.ToString();
}
return output;
}
我使用它,并在基类中使用[XmlInclude]派生类。然后是OmitType和RemoveAllNamespaces。本质上,然后将派生类视为基类。