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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: different first element in a list
on 24-02-2003 10:07, Michael Kay at michael.h.kay@xxxxxxxxxxxx wrote:
> In the inner for-each, instead of
>
> xsl:for-each select="key('k', @type)"
>
> try
>
> xsl:for-each select="key('k', @type)[position()!=1]"
Now it works :) Thank you very much...
but I still have a lot of problems in understanding how this works
Can somebody help me to understand commenting the code?
The Xslt I is at
<http://biografica.tzone.it/cv/eucv_eng_prova.xsl>)
Let's start from the basic concepts:
If I have an xml list like this
__________________________________________
Xml
<xml>
<element>1</element>
<element>3</element>
<element>2</element>
<element>4</element>
<element>5</element>
</xml>
__________________________________________
how can I obtain an ordered list like this
__________________________________________
Xhtml
<table>
<tr>
<td>list:</td>
<td>1</td>
</tr>
<tr>
<td/>
<td>2</td>
</tr>
<tr>
<td/>
<td>3</td>
</tr>
<tr>
<td/>
<td>4</td>
</tr>
<tr>
<td/>
<td>5</td>
</tr>
</table>
__________________________________________
Rendering
list: 1
2
3
4
5
Thank you very much :)
And excuse me if I don't understand, but I have not a programmer's brain (or
I'm simply stupid :P)
If you need the help of an information architect, please ask freely :)
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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