我正在使用该类将 XSL-T 文件xsltUri
应用于 XML 文件:TargetXmlFile
XslCompiledTransform
XslCompiledTransform xslTransform = new XslCompiledTransform(false);
xslTransform.Load(xsltUri);
using (var outStream = new MemoryStream())
{
var writer = new StreamWriter(outStream, new UTF8Encoding());
using (var reader = new XmlTextReader(TargetXmlFileName)
{
WhitespaceHandling = WhitespaceHandling.All,
DtdProcessing = DtdProcessing.Ignore
})
{
xslTransform.Transform(reader, xsltArguments, writer);
}
outStream.Position = 0;
using (FileStream outFile = new FileStream(outputFileName, FileMode.Create))
{
outStream.CopyTo(outFile);
}
}
输入 XML:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<element
id="1"
attr1="value11"
attr2="value12"/>
<element id="2" attr1="value21" attr2="value22"/>
</root>
输入 XSL:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="//element[@id='2']/@attr1">
<xsl:attribute name="attr1">
<xsl:value-of select="'newvalue21'"/>
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
实际输出 XML:
<?xml version="1.0" encoding="utf-8"?><root>
<element id="1" attr1="value11" attr2="value12" />
<element id="2" attr1="newvalue21" attr2="value22" />
</root>
所需的输出 XML:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<element
id="1"
attr1="value11"
attr2="value12"/>
<element id="2" attr1="newvalue21" attr2="value22"/>
</root>
问题:如何在输出 XML 文件的“元素”标签内保留输入 XML 文件的空格(尤其是换行符)?我尝试了不同的选项,但在这种情况下没有任何效果。
感谢您的任何提示!