Subject: Re: [xsl] Re:Problem w/ call-template From: Joerg Heinicke <joerg.heinicke@xxxxxx> Date: Fri, 31 May 2002 00:47:00 +0200 |
Thank you both of you for quickly spotting the problem. I closed the template in the wrong place as Peter pointed out. Nilesh explicitly spelled it out for me. :>)))
I have made the correction; that fixed the problem. However, somehow, w/ the
existing logic, "originalString" doesn't seem to be passed correctly. It
prints out correctly the "String" where it was selected ( before calling the
template lastSubString. However, the result is empty since the
"originalString" is empty in the template "lastSubString". How do I print out
the "$originalString" in the location marked ***DEBUG: for debugging?
Thanks,
Kim
Date: Thu, 30 May 2002 07:41:19 +0000 From: "NILESH PATEL" <jayganesh786@xxxxxxxxxxx> Subject: Re: [xsl] Problem w/ call-template
Hi Kim,
I think even then it will still complain about template name do not match or something. I think your xsl fragment should look like this. Correct me if I am wrong, as I am not really a master in xsl.
<xsl:template match="String"> [...] <xsl:call-template name="lastSubString"> <xsl:with-param name="originalString" select="."/>
***DEBUG1:
</xsl:call-template> </xsl:template>
<xsl:template name="lastSubString"> <xsl:param name="originalString"/>
***DEBUG2:
<xsl:choose> <xsl:when test="contains($originalString,',')"> <xsl:call-template name="lastSubString"> <xsl:with-param name="originalString" select="substring-after($originalString,',')"/> </xsl:call-template> </xsl:when> </xsl:choose> Last substring: <xsl:value-of select="$originalString"/> </xsl:template>
See if this works, which I am preety sure(99.9%) will work. Let me know the resultplease so I can judge myself.
Thanks, kepp the good work going.
Nilesh Patel
From: Peter Davis <pdavis152@xxxxxxxxx> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Problem w/ call-template Date: Wed, 29 May 2002 19:53:13 -0700
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1
On Wednesday 29 May 2002 16:11, Kim wrote:
Hi,
I got an error msg: Could not find template named: "lastSubString". It
is
from the "***" line. What am I missing here? I checked the spelling;
it
is fine.
...
</xsl:template> </xsl:template>
You have defined the "lastSubString" template *within* another template. Really you shouldn't have gotten this far -- <xsl:template> must be a top-level element (that is, a child of <xsl:stylesheet>). Simply move the "lastSubString" template outside of the other template and it should work.
- -- Peter Davis
===== Kim
************************* May the force be with you. *************************
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