在xsl转换后,将<？xml version ="1.0"？>添加到xml文件的顶部

Question

我已经成功地使用xsl模板将带有数据的简单xml文件转换为另一个xml文件(excel模板),这就是我的xsl文件的样子:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
  <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:output method="xml" standalone="yes"/>

    <xsl:template match="/">
      <xsl:processing-instruction name="mso-application">
        <xsl:text>progid="Excel.Sheet"</xsl:text>
      </xsl:processing-instruction>
    ...(stuff here)...
    </xsl:template>
  </xsl:stylesheet>

生成的xml文件正确写出,但包含除外

<?xml version="1.0"?>

在文件的顶部.如何让它出现在顶部？

目前我生成的xml文件以:

<?mso-application progid="Excel.Sheet"?>
...(rest of file)...

但我需要做的是:

<?xml version="1.0"?>
<?mso-application progid="Excel.Sheet"?>
.(rest of file)...

我正在使用以下代码通过Windows窗体进行此转换:

XPathDocument myXPathDoc = new XPathDocument(xmlfile);
XslCompiledTransform myXslTrans = new XslCompiledTransform();
myXslTrans.Load(xslfile);
XmlTextWriter myWriter = new XmlTextWriter(xmlexcelfile, null);
myWriter.Formatting = Formatting.Indented;
myWriter.Namespaces = true;
myXslTrans.Transform(myXPathDoc, null, myWriter);
myWriter.Close();

我试过玩xsl:output standalone="yes/no",以及omit-xml-declaration="no".我还试过(在C#中)代码myWriter.WriteStartDocument();在转换之前添加,但这是不允许的.我也尝试在网上搜索这个并继续回到standalone="yes"但是那不起作用.这里有什么我想念的吗？哦,万一你想知道为什么我需要

<?xml version="1.0"?>

在结果文件的顶部,这是因为当使用excel打开xml文件时,excel无法正确识别它,但如果包含它,则excel会正确打开它...

Answer 1

你可以这样做

<xsl:output method="xml" indent="yes" omit-xml-declaration="no" />

或类似的东西使用 XmlWriterSettings

编辑:添加更多代码.以前缺少一些部分

        XmlWriterSettings writerSettings = null;
        XsltArgumentList transformationArguments = null;
        XslCompiledTransform transformer = null;
        MemoryStream memoryStream = null;
        XPathDocument xPathDocument = null;
        StringBuilder sb = null;
        XmlWriter writer = null;
        XmlDocument resultXml = null;
        try
        {
            writerSettings = new XmlWriterSettings();
            writerSettings.OmitXmlDeclaration = false; // This does it
            writerSettings.Indent = true;

            transformationArguments = new XsltArgumentList();
            transformer = new XslCompiledTransform();
            memoryStream = new MemoryStream(System.Text.Encoding.Default.GetBytes(xml.OuterXml));
            xPathDocument = new XPathDocument(new StreamReader(memoryStream));
            sb = new StringBuilder();
            // give the settings to the writer here
            writer = XmlWriter.Create(sb, writerSettings);
            // this is not mandatory, obviously, just passing parameters to my xslt file
            foreach (KeyValuePair<string, object> parameter in parameters)
            {
                transformationArguments.AddParam(parameter.Key, string.Empty, parameter.Value);
            }

            using (Stream strm = Assembly.GetExecutingAssembly().GetManifestResourceStream("Lib.XSLTFile1.xslt"))
            using (XmlReader reader = XmlReader.Create(strm))
            {
                transformer.Load(reader);
                transformer.Transform(xPathDocument, transformationArguments, writer);
            }
            resultXml = new XmlDocument();
            resultXml.LoadXml(sb.ToString());
            // for testing only
            File.AppendAllText(@"Your path goes here\result.xml", resultXml.OuterXml);
        }
        catch (Exception)
        {
            throw;
        }

这就是我这样做的方法,但是这段代码专门用于创建一个实例XmlDocument.我相信你能适应你的需求.