Re: [xsl] How do I change a XSL style sheet to group data together under one heading

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
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.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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