|
Subject: [xsl] Problems grouping adjacent siblings From: "Huditsch Roman" <Roman.Huditsch@xxxxxxxxxxxxx> Date: Thu, 30 Jun 2005 08:37:54 +0200 |
Good morning list,
I would like to group adjacent siblings which share the same local-name
into a container element.
Given this structure
<ax>
<e></e>
<r></r>
<exp>Example</exp>
<exp>bla bla</exp>
<exp>jada jada</exp>
<b></b>
<e></e>
<exp>Example 2</exp>
<exp>bla bla</exp>
<v></v>
</ax>
<bx>
<c></c>
<exp>Example 3</exp>
<exp>bla bla</exp>
<b></b>
</bx>
The following output should be produced:
<ax>
<e></e>
<r></r>
<remark>
<exp>Example</exp>
<exp>bla bla</exp>
<exp>jada jada</exp>
</remark>
<b></b>
<e></e>
<remark>
<exp>Example 2</exp>
<exp>bla bla</exp>
</remark>
<v></v>
</ax>
<bx>
<c></c>
<r></r>
<remark>
<exp>Example 3</exp>
<exp>bla bla</exp>
</remark>
<b></b>
</bx>
I thought that this would be the classical use-case for an
<xsl:for-each-group> with @group-adjacent, but I am getting the
strangest results...
I don't even know if it is better to do the grouping in a template
matching
exp[not(preceding-sibling::*[1][local-name()='exp']) or in a template
matching exp's parent
I also tried to find a @group-starting-with solution, since the first
<exp> always starts with the string 'Example'. But with no success
either....
Can you help me with this problem?
Thank you very much for your help!
Wbr,
Roman
| Current Thread |
|---|
|
| <- Previous | Index | Next -> |
|---|---|---|
| Re: [xsl] sorting issue, omprakash . v | Thread | Re: [xsl] Problems grouping adjacen, Mukul Gandhi |
| Re: [xsl] sorting issue, omprakash . v | Date | Re: [xsl] Problems grouping adjacen, Mukul Gandhi |
| Month |