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I have a xslt page that looks like this
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlnssl="http://www.w3.org/1999/XSL/Transform"
xmlns:lc="http://mark/DevelopmentWebsite">
<xsl:template match="/">
<h2>My Game Collection</h2>
<table border="1">
<xsl:for-each select="lc:Entertainment/lc:GameList">
<tr>
<th colspan="2"><xsl:value-of select="lc:Title"/></th>
</tr>
<tr>
<td>
<xsl:element name="IMG">
<xsl:attribute name="SRC">
<xsl:value-of select="lc:Image"/>
</xsl:attribute>
</xsl:element>
</td>
<td><xsl:value-of select="lcescription"/></td>
</tr>
</xsl:for-each>
</table>
</xsl:template>
</xsl:stylesheet>

All I would have to do to get it to display all games that are in the
Horror genre is put

<xsl:for-each select="lc:Entertainment/lc:GameList[lc:Genre =
'Horror']">

instead of

<xsl:for-each select="lc:Entertainment/lc:GameList">

Now how could I get it so there is a drop down list on my page with a
list of different genres. If the person selects a certain genre it
passes the argument to the XSLT so it only displays the genre the user
picked? Can I even pass arguments to a XSLT page?

> Can I even pass arguments to a XSLT page?

An XSLT transformation can be passed an external parameter. How this is
accomplished depends on the particular vendor. For example in the case of
MSXML one can use the IXSLProcessor.addParameter() method.


=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL