定義如下類
[XmlRoot("funderService")]
public class Request_gl
{
public Request_gl()
{
ParamList = new List<string>();
}
[XmlAttribute]
public string functionName { get; set; }
[XmlArrayItem("vaule")]
public List<string> ParamList { get; set; }
public string beanToXml()
{
//準備序化列物件
var format = new XmlSerializer(typeof(Request_gl));
var stringBuilder = new StringBuilder();
//設定序序化XML格式
XmlWriterSettings xws = new XmlWriterSettings();
xws.Indent = true;
xws.OmitXmlDeclaration = true;
xws.Encoding = Encoding.UTF8;
XmlWriter xmlWriter = XmlTextWriter.Create(stringBuilder, xws);
//去掉要結點的 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" 屬性
XmlSerializerNamespaces _namespaces = new XmlSerializerNamespaces(
new XmlQualifiedName[] {
new XmlQualifiedName(string.Empty, string.Empty)
});
xmlWriter.WriteStartDocument(true);
format.Serialize(xmlWriter, this, _namespaces);
return stringBuilder.ToString();
}
}
需要序列化成的XML格式如下
<funderService functionName='2023'>
<value>1</value>
<value>2</value>
<value>3</value>
</funderService >
可實際序列化出來是這樣子的
<funderService functionName="2023">
<ParamList>
<vaule>1</vaule>
<vaule>2</vaule>
<vaule>3</vaule>
<vaule>4</vaule>
</ParamList>
</funderService>
多了 <ParamList> 節點
請問型別該如何定義才能序列化成想要的
uj5u.com熱心網友回復：
[XmlArrayItem("vaule")]
//替換為
[XmlElement("vaule")]
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標籤：C#
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