Question

我使用此代码创建了一个使用XMLWriter创建的简单XML文件：

using (XmlWriter writer = XmlWriter.Create("c:\\temp\\data.xml"))
{

     writer.WriteStartElement("ScaleFactors");  

     writer.WriteStartAttribute("CorrectionFactorX");
     writer.WriteValue(CorrectionFactorX);
     writer.WriteEndAttribute();

     writer.WriteStartAttribute("CorrectionFactorY");
     writer.WriteValue(CorrectionFactorY);
     writer.WriteEndAttribute();

     writer.WriteStartAttribute("CorrectionFactorZ");
     writer.WriteValue(CorrectionFactorZ);
     writer.WriteEndAttribute();

     writer.WriteEndElement();

     writer.Flush();
}

XML文件看起来像这样：

<?xml version="1.0" encoding="utf-8"?>
    <ScaleFactors CorrectionFactorX="1" 
                  CorrectionFactorY="1" 
                  CorrectionFactorZ="1" />

（CorrectionFactorX / Y / Z是双打）现在我想用XMLReader读回它们但是失败了。我试过MoveToAttribute / GetAttribute没有成功。任何提示和代码示例都表示赞赏。

谢谢！

Answer 1

如果你可以使用XDoc，那么

void Main()
{
    var xml = "<?xml version=\"1.0\" encoding=\"utf-8\"?><ScaleFactors CorrectionFactorX=\"1.31\" CorrectionFactorY=\"1\" CorrectionFactorZ=\"1\" />";
    var xdoc = XDocument.Parse(xml);

    var cfx = Double.Parse(xdoc.Element("ScaleFactors").Attribute("CorrectionFactorX").Value);
    Console.WriteLine (cfx);    
}

如果你坚持使用XmlReader，那么

void Main()
        {
            var xml = "<?xml version=\"1.0\" encoding=\"utf-8\"?><ScaleFactors CorrectionFactorX=\"1.31\" CorrectionFactorY=\"1\" CorrectionFactorZ=\"1\" />";
            using (var reader = XmlReader.Create(new StringReader(xml)))
            {
                while (reader.Read())
                {
                    if (reader.NodeType == XmlNodeType.Element)
                    {
                        Console.WriteLine(reader.Name);
                        var attc = reader.AttributeCount;
                        for (int i = 0; i < attc; i++)
                        {
                            Console.WriteLine(reader.GetAttribute(i));
                        }
                    }
                }
            }
        }

（MSDN处理XML不同部分的好例子，但XDocument更好“

Answer 2

使用XmlDocument -

        var xml = new XmlDocument();
        xml.Load("xmlPath");

        var match = xml.DocumentElement.Attributes;
        foreach (XmlAttribute item in match)
        {
            Console.WriteLine(item.Value);
        }

使用XMLReader从xml文件中读取double

2 个答案: