我有一个方案,其中CDATA返回html标记,所以我必须使用disable-output-escaping =“ yes”才能将内容正确呈现到html页面。但是,CDATA也可能包含&字符,这导致我的页面无法通过w3c验证。
这是我的xml来源的示例:
<?xml version="1.0" encoding="UTF-8"?>
<jobs>
<job>
<positionTitle><![CDATA[Health & Safety Officer]]></positionTitle>
<description1><![CDATA[<p>You must have experience of working in a health & safety team.</p>]]></description1>
</job>
</jobs>
XSLT
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:content="http://purl.org/rss/1.0/modules/content/" >
<xsl:output method="html"/>
<xsl:template match="/">
<xsl:for-each select="jobs/job">
<div class="job">
<div class="title"><h3><xsl:value-of select="positionTitle"/></h3></div>
<div class="description"><xsl:value-of select="description1" disable-output-escaping="yes"/></div>
</div>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
结果:
<div class="title">
<h3>Health & Safety Officer</h3>
</div>
<div class="description">
<p>You must have experience of working in a health & safety team.</p>
</div>
description元素中的&字符验证失败,如何将&
转换为&
?
谢谢
答案 0 :(得分:1)
在 XSLT 2.0 中,您可以使用replace()
函数来逃离&符:
<xsl:value-of select="replace(description1, '&', '&amp;')" disable-output-escaping="yes"/>
在 XSLT 1.0 中,您需要使用递归模板:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html"/>
<xsl:template match="/">
<xsl:for-each select="jobs/job">
<div class="job">
<div class="title">
<h3>
<xsl:value-of select="positionTitle"/>
</h3>
</div>
<div class="description">
<xsl:call-template name="replace">
<xsl:with-param name="text" select="description1"/>
</xsl:call-template>
</div>
</div>
</xsl:for-each>
</xsl:template>
<xsl:template name="replace">
<xsl:param name="text"/>
<xsl:param name="searchString">&</xsl:param>
<xsl:param name="replaceString">&amp;</xsl:param>
<xsl:choose>
<xsl:when test="contains($text,$searchString)">
<xsl:value-of select="substring-before($text,$searchString)" disable-output-escaping="yes"/>
<xsl:value-of select="$replaceString" disable-output-escaping="yes"/>
<!-- recursive call -->
<xsl:call-template name="replace">
<xsl:with-param name="text" select="substring-after($text,$searchString)"/>
<xsl:with-param name="searchString" select="$searchString"/>
<xsl:with-param name="replaceString" select="$replaceString"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$text" disable-output-escaping="yes"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
答案 1 :(得分:0)
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">
<xsl:character-map name="a">
<xsl:output-character character="&" string="&amp;"/>
<xsl:output-character character="<" string="<"/>
<xsl:output-character character=">" string=">"/>
</xsl:character-map>
<xsl:output method="html" use-character-maps="a"/>
<xsl:template match="/">
<xsl:for-each select="jobs/job">
<div class="job">
<div class="title"><h3><xsl:value-of select="positionTitle"/></h3></div>
<div class="description"><xsl:value-of select="description1"/></div>
</div>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
You may also use character map.