Question

我有一个样本xml文件，看起来像这样。

    <Window x:Class="WpfApplication1.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    Title="MainWindow" Height="350" Width="525">
<Grid>
    <StackPanel>
        <TextBlock Text="First Text" Margin="5"/>
        <Label Content="Second Text" HorizontalAlignment="Center"/>
        <TextBox Text="Third Text"/>
        <GroupBox Header="Fourth Text">
            Fifth Text
            that extends to another line.
        </GroupBox>
        <Button Content="Sixth Text"/>
        <Frame Content="&lt;Seventh Text&gt;"></Frame>
        <ComboBox>
            Eighth Text</ComboBox>
        <Label Content="{Binding LabelText}" HorizontalAlignment="Center"/>
    </StackPanel>
</Grid>

我的输出文件如下所示：

    (4)Title="MainWindow"
    (7)Text="First Text"
    (8)Content="Second Text"
    (9)Text="Third Text"
    (10)Header="Fourth Text"
    (11) Fifth Text                that extends to another line.
    (14)Content="Sixth Text"
    (15)Content="&lt;Seventh Text&gt;"
    (17) Eighth Text

这主要是我想要的。但是，出于某种原因，我只能获得＆＃34; Title＆＃34;和＆＃34;文字＆＃34;和＆＃34;内容＆＃34;等等。但我希望它打印出来＃34; TextBlock Text＆＃34;和＆＃34;标签内容＆＃34;和＆＃34; TextBox Text＆＃34;和＆＃34;按钮内容＆＃34;等等。我正在使用XmlTextReader，但我似乎无法找到任何支持打印出来的。 reader.Name只是打印出我已经拥有的东西这是我的代码：

    public void ParseXml(String filename)
    {
        XmlTextReader reader = new XmlTextReader(filename);
        while (reader.Read())
        {
            switch (reader.NodeType)
            {
                case XmlNodeType.Element:
                    for (int i = 0; i < reader.AttributeCount; i++)
                    {
                        reader.MoveToAttribute(i);
                        if (reader.LineNumber < 4)
                        {
                            continue;
                        }
                        //WriteLine(Path.GetFullPath(filename));
                        if(reader.Name != "Width" && reader.Name != "Height" && reader.Name != "Margin"
                            && reader.Name != "HorizontalAlignment")
                            WriteLine("(" + reader.LineNumber + ")" + reader.ReadOuterXml());                         
                    }
                    break;
                case XmlNodeType.Text:
                    WriteLine("(" + (reader.LineNumber + 1) + ") " + reader.Value.Replace("\r\n","").Trim());
                    break;
                case XmlNodeType.EndElement:
                    break;
            }
        }
        reader.Close();
    }

谢谢！

Answer 1

正如我在评论中暗示的那样，当读者仍然在元素上时，你需要得到reader.Name的结果 - 之前你通过调用reader.MoveToAttribute(i)来移动它。

除非您有特殊原因（例如性能或非常大的文件），否则使用更高级别的API（如LINQ to XML）会更容易。

一个简单的实现可能如下所示：

var document = XDocument.Load(filename, LoadOptions.SetLineInfo);

var excludedAttributes = new HashSet<XName>
{
    "{http://schemas.microsoft.com/winfx/2006/xaml}Class",
    "Width",
    "Height",
    "Margin",
    "HorizontalAlignment"
};

foreach (var element in document.Descendants())
{
    IXmlLineInfo lineInfo = element;

    var sb = new StringBuilder();

    sb.AppendFormat("({0}) ", lineInfo.LineNumber);
    sb.Append(element.Name.LocalName);

    var outputAttributes = element.Attributes()
        .Where(a => !a.IsNamespaceDeclaration && !excludedAttributes.Contains(a.Name));

    foreach (var attribute in outputAttributes)
    {
        sb.AppendFormat(" {0}=\"{1}\"", attribute.Name, attribute.Value);
    }

    if (element.Nodes().OfType<XText>().Any())                
    {
        sb.Append(" ");
        sb.Append(element.Value.Replace("\n", string.Empty).Trim());
    }

    WriteLine(sb.ToString());
}

如何获取Xml节点的全名

1 个答案: