我正在尝试获取此xml文件主节点的logicalName属性的值:
<?xml version="1.0" encoding="ISO-8859-1"?>
<ticketlayout xmlns="http://www.example.com/ticketlayout" logicalName="target.xml" deviceCode="1" measurement="mm" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.com/ticketlayout">
<fontdefinition id="BarCode">
<fontname>Code128bWin</fontname>
<size measure="pt">16</size>
</fontdefinition>
</ticketlayout>
我尝试通过以下方式添加名称空间“ xsi”,“ http://www.w3.org/2001/XMLSchema-instance”:
XmlDocument fLayout = new XmlDocument();
fLayout.Load("myFile.xml");
XmlNamespaceManager nsmRequestLayout = new XmlNamespaceManager(fLayout.NameTable);
nsmRequestLayout.AddNamespace("xsi", "http://www.w3.org/2001/XMLSchema-instance");
string sValue = fLayout.SelectNodes("//ticketlayout", nsmRequestLayout)[0].Attributes["name"].Value;
但是我没有节点。我试过没有命名空间,再也没有节点,然后继续。 ¿可以请任何人帮我吗?
谢谢。
答案 0 :(得分:1)
如果要获取值:target.xml 试试这个代码
XmlDocument fLayout = new XmlDocument();
fLayout.Load("myFile.xml"); // your XML file
var attrib = fLayout["ticketlayout"].Attributes["logicalName"].Value;
答案 1 :(得分:1)
首先,您的XML无效。 我进行了修改,以实现所需的目的。 XML文件:
<?xml version="1.0" encoding="UTF-8"?>
<ticketlayout xmlns="http://www.example.com/ticketlayout" logicalName="target.xml" deviceCode="1" measurement="mm" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.com/ticketlayout">
<fontdefinition id="BarCode">
<fontname>Code128bWin</fontname>
<size measure="pt">16</size>
</fontdefinition>
</ticketlayout>
我不确定为什么没有模型结构来反序列化xml,然后访问所需的任何属性/属性。
示例: 班级:
[XmlRoot(ElementName = "size", Namespace = "http://www.example.com/ticketlayout")]
public class Size
{
[XmlAttribute(AttributeName = "measure")]
public string Measure { get; set; }
[XmlText]
public string Text { get; set; }
}
[XmlRoot(ElementName = "fontdefinition", Namespace = "http://www.example.com/ticketlayout")]
public class Fontdefinition
{
[XmlElement(ElementName = "fontname", Namespace = "http://www.example.com/ticketlayout")]
public string Fontname { get; set; }
[XmlElement(ElementName = "size", Namespace = "http://www.example.com/ticketlayout")]
public Size Size { get; set; }
[XmlAttribute(AttributeName = "id")]
public string Id { get; set; }
}
[XmlRoot(ElementName = "ticketlayout", Namespace = "http://www.example.com/ticketlayout")]
public class Ticketlayout
{
[XmlElement(ElementName = "fontdefinition", Namespace = "http://www.example.com/ticketlayout")]
public Fontdefinition Fontdefinition { get; set; }
[XmlAttribute(AttributeName = "xmlns")]
public string Xmlns { get; set; }
[XmlAttribute(AttributeName = "logicalName")]
public string LogicalName { get; set; }
[XmlAttribute(AttributeName = "deviceCode")]
public string DeviceCode { get; set; }
[XmlAttribute(AttributeName = "measurement")]
public string Measurement { get; set; }
[XmlAttribute(AttributeName = "xsi", Namespace = "http://www.w3.org/2000/xmlns/")]
public string Xsi { get; set; }
[XmlAttribute(AttributeName = "schemaLocation", Namespace = "http://www.w3.org/2001/XMLSchema-instance")]
public string SchemaLocation { get; set; }
}
然后您可以使用序列化程序:
public class Serializer
{
public T Deserialize<T>(string input) where T : class
{
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
using (StringReader stringReader = new StringReader(input))
{
return (T)xmlSerializer.Deserialize(stringReader);
}
}
public string Serialize<T>(T ObjectToSerialize)
{
XmlSerializer xmlSerializer = new XmlSerializer(ObjectToSerialize.GetType());
StringBuilder builder = new StringBuilder();
using (StringWriterWithEncoding textWriter = new StringWriterWithEncoding(builder, Encoding.UTF8))
{
xmlSerializer.Serialize(textWriter, ObjectToSerialize);
return textWriter.ToString();
}
}
}
public class StringWriterWithEncoding : StringWriter
{
Encoding encoding;
public StringWriterWithEncoding(StringBuilder builder, Encoding encoding)
: base(builder)
{
this.encoding = encoding;
}
public override Encoding Encoding
{
get { return encoding; }
}
}
最后,通过执行以下操作,您可以访问所需的任何内容:
var serializer = new Serializer();
//I used a local file for testing, but it should be the same thing with your file
var xmlInputData = File.ReadAllText(@"MyXmlPath");
var output = serializer.Deserialize<Ticketlayout >(xmlInputData);
var logicalName = output.LogicalName;