Subject: Re: [xsl] How do I change a XSL style sheet to group data together under one heading From: "kieters c" <kieters@xxxxxxxxxxx> Date: Mon, 21 May 2007 10:41:21 +0000 |
From: Abel Braaksma <abel.online@xxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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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