[xsl] filter based on parameter, not getting it.

["Carter, Will" <WCarter@xxxxxxxxxxxxxxxx>]

Hi,

I am trying to filter based on a parameter.  If the filterparam is empty, don't do any filtering.  If filterparam is something, that's what to filter on.

xml.xml
---------------------------------------------------------
<people>
	<person name="george">
		<cat>zoro</cat>
		<dog>butch</dog>
		<fish>jaws</fish>
	</person>
	<person name="jennifer">
		<cat>felix</cat>
		<dog>fido</dog>
		<fish>moby</fish>
	</person>
	<person name="simon">
		<cat>tom</cat>
		<dog>scooby</dog>
		<fish>conroy</fish>
	</person>
</people>
---------------------------------------------------------


This code doesn't work:
---------------------------------------------------------
<xsl:param name="filterparam"/>
<xsl:template match="/">

<xsl:choose>
	<xsl:when test="string-length($filterparam) = 0">
		<xsl:variable name="filter" select="//person"/>
	</xsl:when>
	<xsl:otherwise>
		<xsl:variable name="filter" select="//person[cat = $filterparam]"/>
	</xsl:otherwise>
</xsl:choose>

<table>
	<xsl:for-each select="$filter">
	<tr>
		<td><xsl:value-of select="cat"/></td>
		<td><xsl:value-of select="dog"/></td>
		<td><xsl:value-of select="fish"/></td>
	</tr>
	</xsl:for-each>
</table>
---------------------------------------------------------

here is my command line entry:
java org.apache.xalan.xslt.Process -in xml.xml -xsl filterprob.xsl -param filterparam zoro -out output.html

my error:
Could not find variable with the name of filter

I wanted my output to be:
zoro butch jaws

I realize that the error is caused because the variable is set up in the choose. if I move the variable statements outside the choose, it works...but then I am not checking if filterparam is empty or not.

thanks for any help you could provide.

will

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