Re: [xsl] different first element in a list

[Lorenzo De Tomasi <detomasi.liste@xxxxxxxxx>]

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 :)


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