在没有HTML编码最终输出的情况下，使用XSL进行转换的属性方法是什么? [英] What's the property way to transform with XSL without HTML encoding my final output?

问题描述

因此，我正在使用.NET.我有一个XSL文件，即C#中的XslTransform对象，它读取XSL文件并将一个XML数据(内部制造)转换为HTML.

So, I am working with .NET. I have an XSL file, XslTransform object in C# that reads in the XSL file and transforms a piece of XML data (manufactured in-house) into HTML.

我注意到我的最终输出已自动编码为< 和& gt;的< 和> ..有什么办法可以防止这种情况发生?有时我需要加粗或斜体显示文本，但无意中将其清除了.

I notice that my final output has < and > automatically encoded into &lt; and &gt;. Is there any ways I can prevent that from happening? Sometimes I need to bold or italicize my text but it's been unintentionally sanitized.

推荐答案

您的xsl文件应具有:

Your xsl file should have:

• html的输出
• 省略xslt中使用的所有名称空间

即

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt"
    exclude-result-prefixes="xsl msxsl">

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

    <!-- lots -->
</xsl:stylesheet>

并且理想情况下，您应该使用接受 TextWriter 或 Stream (不是 XmlWriter )的重载-例如:

And you should ideally use the overloads that accept either a TextWriter or a Stream (not XmlWriter) - i.e. something like:

StringBuilder sb = new StringBuilder();
using (XmlReader reader = XmlReader.Create(source)
using (TextWriter writer = new StringWriter(sb))
{
    XslCompiledTransform xslt = new XslCompiledTransform();
    xslt.Load("Foo.xslt"); // in reality, you'd want to cache this
    xslt.Transform(reader, options.XsltOptions, writer);
}
string html = sb.ToString();

在xslt中，如果您确实想要独立的< /> (即，由于某种原因而希望其格式错误)，则需要禁用输出转义:

In the xslt, if you really want standalone < / > (i.e. you want it to be malformed for some reason), then you need to disable output escaping:

<xsl:text disable-output-escaping="yes">
    Your malformed text here
</xsl:text>

但是，通常转义字符是正确.

However, in general it is correct to escape the characters.

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