Subject: Re: How do I change a XSL style sheet to group data together under one heading
From: Abel Braaksma <abel.online@xxxxxxxxx>
Date: Mon, 21 May 2007 10:35:42 +0200
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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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