[英]Serialize type into XML from .NET
I have this C# 4.0 type 我有这个C#4.0类型
public class DecimalField
{
public decimal Value { get; set; }
public bool Estimate { get; set; }
}
I want to use XmlSerializer to serialize the type into 我想使用XmlSerializer将类型序列化为
<Val Estimate="true">123</Val>
Ideally, I want to omit the Estimate attribute if its value is false. 理想情况下,如果其值为false,我想省略Estimate属性。 Changing Estimate to a nullable bool is acceptable. 将估计值更改为可以为空的bool是可以接受的。
What attributes/implementations are required to go from this type to this XML representation? 从此类型到此XML表示需要哪些属性/实现?
Thanks. 谢谢。
Conditionally omitting Estimate
would require a lof of coding. 有条件地省略Estimate
将需要自我编码。 I wouldn't go that way. 我不会这样。
XmlWriter writer = XmlWriter.Create(stream, new XmlWriterSettings() { OmitXmlDeclaration = true });
var ns = new XmlSerializerNamespaces();
ns.Add("", "");
XmlSerializer xml = new XmlSerializer(typeof(DecimalField));
xml.Serialize(writer, obj, ns);
- -
[XmlRoot("Val")]
public class DecimalField
{
[XmlText]
public decimal Value { get; set; }
[XmlAttribute]
public bool Estimate { get; set; }
}
You can also manually serialize your class using Linq2Xml 您也可以使用Linq2Xml手动序列化您的类
List<XObject> list = new List<XObject>();
list.Add(new XText(obj.Value.ToString()));
if (obj.Estimate) list.Add(new XAttribute("Estimate", obj.Estimate));
XElement xElem = new XElement("Val", list.ToArray());
xElem.Save(stream);
This is about as close as you can get (an Estimate attribute is always included) without implementing IXmlSerializable: 这是尽可能接近(不包括Estimate属性)而不实现IXmlSerializable:
[XmlRoot("Val")]
public class DecimalField
{
[XmlText()]
public decimal Value { get; set; }
[XmlAttribute("Estimate")]
public bool Estimate { get; set; }
}
With IXmlSerializable, your class looks like this: 使用IXmlSerializable,您的类看起来像这样:
[XmlRoot("Val")]
public class DecimalField : IXmlSerializable
{
public decimal Value { get; set; }
public bool Estimate { get; set; }
public void WriteXml(XmlWriter writer)
{
if (Estimate == true)
{
writer.WriteAttributeString("Estimate", Estimate.ToString());
}
writer.WriteString(Value.ToString());
}
public void ReadXml(XmlReader reader)
{
if (reader.MoveToAttribute("Estimate") && reader.ReadAttributeValue())
{
Estimate = bool.Parse(reader.Value);
}
else
{
Estimate = false;
}
reader.MoveToElement();
Value = reader.ReadElementContentAsDecimal();
}
public XmlSchema GetSchema()
{
return null;
}
}
You can test your class like this: 您可以像这样测试您的课程:
XmlSerializer xs = new XmlSerializer(typeof(DecimalField));
string serializedXml = null;
using (StringWriter sw = new StringWriter())
{
DecimalField df = new DecimalField() { Value = 12.0M, Estimate = false };
xs.Serialize(sw, df);
serializedXml = sw.ToString();
}
Console.WriteLine(serializedXml);
using (StringReader sr = new StringReader(serializedXml))
{
DecimalField df = (DecimalField)xs.Deserialize(sr);
Console.WriteLine(df.Estimate);
Console.WriteLine(df.Value);
}
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