Reputation: 9301
If I select a DIV tag in a XHTML file with XSLT, like //*[@id='background']
How do I add a style, like a background color or other CSS styles like borders to the DIV? And if I have a list inside the DIV ID=background, how can I style the list, like removing the bullets? :)
Upvotes: 3
Views: 10379
Reputation: 24816
This is really easy with XSLT. For instance, your input is:
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
<title></title>
</head>
<body>
<div id="background">
<ul style="list-style-type: bullet">
<li>Coffee</li>
<li>Tea</li>
<li>Milk</li>
</ul>
</div>
</body>
</html>
You can use the identity transform to copy the input XML as is, and override the nodes of interest:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns="http://www.w3.org/1999/xhtml"
xmlns:x="http://www.w3.org/1999/xhtml"
exclude-result-prefixes="x">
<xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*" />
</xsl:copy>
</xsl:template>
<xsl:template match="x:div[@id='background']">
<xsl:copy>
<xsl:attribute name="style">
<xsl:text>border-style:solid;border-width:medium</xsl:text>
</xsl:attribute>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="x:ul[ancestor::*
[name()='div' and @id='background']]/@style">
<xsl:attribute name="style">
<xsl:text>list-style-type: none</xsl:text>
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
The output will be:
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
<title></title>
</head>
<body>
<div style="border-style:solid;border-width:medium" id="background">
<ul style="list-style-type: none">
<li>Coffee</li>
<li>Tea</li>
<li>Milk</li>
</ul>
</div>
</body>
</html>
Upvotes: 5