Changing class on <P> or <DIV> with XSLT? - xslt

Question

I need to turn <P class="brodtext"> into <P class="normal"> using XSLT.
I've started with:
<xsl:template match="//p[#class='brodtext']">
</xsl:template>
Can you please advise on how to do the change?
Edit:
I need to integrate it into this:
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:proxy="java:senselogic.sitevision.portlet.proxy.web.ProxyFunctions"
extension-element-prefixes="proxy">
<xsl:import href="template.xsl"/>
<!-- H1-H6 needs newline -->
<xsl:template match="h1">
<xsl:copy>
<xsl:attribute name="style">display:inline</xsl:attribute>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="h2">
<xsl:copy>
<xsl:attribute name="style">display:inline</xsl:attribute>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="h3">
<xsl:copy>
<xsl:attribute name="style">display:inline</xsl:attribute>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="h4">
<xsl:copy>
<xsl:attribute name="style">display:inline</xsl:attribute>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="h5">
<xsl:copy>
<xsl:attribute name="style">display:inline</xsl:attribute>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="h6">
<xsl:copy>
<xsl:attribute name="style">display:inline</xsl:attribute>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="//#shape"/>
<xsl:template match="//#clear"/>
<xsl:output method="xml" indent="yes"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="#class[.='brodtext' and parent::P]">
<xsl:attribute name="class">normal</xsl:attribute>
</xsl:template>
<!-- Visa -->
<!-- <xsl:template match="//div[#id='svid10_44e5251f1752aa6808f1edfa']"/> -->
<!-- Dölj -->
<xsl:template match="//div[#id='svid10_44e5251f1752aa6808f1edfb']"/>
<xsl:template match="/" >
<xsl:apply-templates select="//div[#id='svid10_44e5251f1752aa6808f1edf8']" />
</xsl:template>
</xsl:stylesheet>
Hmm... too much code to publish? I guess I have to write something else then.
That's odd.
Man, I don't know what to do, I'm sorry for spamming.
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Answer 1

You could do it simply like this :
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="#class[.='brodtext' and parent::P]">
<xsl:attribute name="class">normal</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
See it working here : https://xsltfiddle.liberty-development.net/93nwMo5/1