Thank you. I did not see the one although I tried to check on all the
brackets. I found another one missing two lines down.
I used XML Notepad 2007 to do the transformation and the result remain the
same as previously reported. Only sample_date_time is seperated and all
other is include in result. I added all the other that should be grouped
with sample_date_time but it made not difference.
Thank you.
Hennie
From: Abel Braaksma <abel(_dot_)online(_at_)xs4all(_dot_)nl>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] How do I change a XSL style sheet to group data together
under one heading
Date: Mon, 21 May 2007 10:35:42 +0200
kieters c wrote:
Good day,
Thank you for your reply. I am aware of the one entry and have tried to
run it with it as presented to you and without it with only
sample_date_time. The result is the same. I receive the message for the
following line:
<xsl:for-each select="sample[generate-id() =
generate-id(key('sample',concat(sample_date_time, cp_name)[1])]">
line 15, Character 23
Stylesheet error: Invalid XPath expression
select
Well, I count three opening parentheses and two closing parentheses so the
processor is correct you have a typo. You do not close the concat()
function:
<xsl:for-each select="sample[generate-id() =
generate-id(key('sample',concat(sample_date_time, cp_name))[1])]">
Cheers,
-- Abel
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