xsl variables

[sb1]

Hi

I am new to xml and I am trying to use xsl to transform my xml
document to smil.

I want to create a line such as
<video src="../media/countries_video/cuba_100.ram "
id="../media/countries_video/greece_100.ram " region="x"/>

for a number of different video files, the references to which are
held within my xml files. At the minute my code:

<xsl:for-each select="$destination/Country/CountryVideo">

<xsl:variable name="countryvid"
select="$destination/Country/CountryVideo"/>
<video src="{$countryvid}" id="{$countryvid}" region="x" />
</xsl:for-each>

produces the correct number of lines but each contains the reference
to the same file, I'm guessing this is because I can't use the one
variable to reference each. I'd appreciate it if anyone could give me
some pointers on how to get around this. (Bearing in mind I'm very new
to all this!)
Thanks in advance

 

[Martin Honnen]

Hi

I am new to xml and I am trying to use xsl to transform my xml
document to smil.

I want to create a line such as
<video src="../media/countries_video/cuba_100.ram "
id="../media/countries_video/greece_100.ram " region="x"/>

for a number of different video files, the references to which are
held within my xml files. At the minute my code:

<xsl:for-each select="$destination/Country/CountryVideo">

<xsl:variable name="countryvid"
select="$destination/Country/CountryVideo"/>
Try

<video src="{$countryvid}" id="{$countryvid}" region="x" />
</xsl:for-each>

produces the correct number of lines but each contains the reference
to the same file, I'm guessing this is because I can't use the one
variable to reference each. I'd appreciate it if anyone could give me
some pointers on how to get around this. (Bearing in mind I'm very new
to all this!)
Thanks in advance

 

[sb1]

Thanks