[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
Re: [xsl] arguments for xsl:call-template
Hi Charly!
I don't understand quite well what you want, because you can do this:
<xsl:with-param name="path" select="'/report/histo/bar'" />
but you can do this ,
<xsl:with-param name="path" select="'/report/histo/@bar'" />
because you are not assigning anything to var. path and also you are doing
histogram just once so the 'for-each' has no reason to be. Explain what's
the purpose of this template.
At 09:38 07/12/01 -0800, you wrote:
Hi everyone.
I'm a beginner with XSL .
Does anyone of you has any idea how to call a template and passing with a
node as argument.
I'm trying the following .
<xsl:template name="histogram">
<xsl:param name="path" />
<xsl:for-each select="$path">
<xsl:value-of select="@value"/>
</xsl:for-each>
</xsl:template>
<xsl:call-template name="histogram">
<xsl:with-param name="path" select="'/report/histo/bar'" />
</xsl:call-template>
my XML looks like .
<report>
<histo>
<bar value="20" />
<bar value="30" />
<bar value="40" />
<bar value="50" />
</histo>
</report>
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
| 
Mac OS X
Linux/Unix
data server software 