xslt / x-path problem

[Bart]

I'm trying to transform an xml file to another xml file by using xslt.
for every child node, i need to do some transformations, except for the
first one, so I can't use a for-each construction.
Does anyone now the correct x-path expression or another way to work around
this problem?
You can find an example of the xml file below
I also have another question, can give pass an argument to an xslt script?

tia
Bart


<root>
<child>
<!--some tags-->
</child>
<child>
<a></a>
</child>
<child>
<a></a>
</child>
<child>
<a></a>
</child>
.....

 

[Joris Gillis]

Hallo Bart,

I'm trying to transform an xml file to another xml file by using xslt.
for every child node, i need to do some transformations, except for the
first one, so I can't use a for-each construction.
Does anyone now the correct x-path expression or another way to work around
this problem?

The problem could be solved with this:

<xsl:for-each select="child[position() &gt; 1]" />

or you could use the push style:
Instead of using for-each, you would then use templates.

<xsl:template match="child">
Some transformation
</xsl:template>

<xsl:template match="root/child[1]"/>

The last template would make sure that the first child is ignored.

regards,

 

[Bart]

Hallo Bart,

Tempore 13:44:57, die Sunday 04 September 2005 AD, hinc in foro

I'm trying to transform an xml file to another xml file by using xslt.
for every child node, i need to do some transformations, except for the
first one, so I can't use a for-each construction.
Does anyone now the correct x-path expression or another way to work
around
this problem?

The problem could be solved with this:

<xsl:for-each select="child[position() &gt; 1]" />

or you could use the push style:
Instead of using for-each, you would then use templates.

<xsl:template match="child">
Some transformation
</xsl:template>

<xsl:template match="root/child[1]"/>

The last template would make sure that the first child is ignored.

regards,

Thanks!

Bart