Re: [xsl] document() loops

["Laura Jenkins" <xsl_list at hotmail dot com>]

Actually the question in the previous thread was a bit different.and i was 
able to solve it.The previous case was i was using one of the XMLs as main 
XML and passing the other xml files as parameters.. that was working fine. 
But then the decision changed to have one xml file with all the xml file 
locations as child elements like..
**** mainxml.xml****
<?xml version="1.0"?>
<!-- all the below xml files( univ1.xml.. etc are in the same directory as 
mainxml.xml
-->
<univ-xml-list>
<univ-xml>univ1.xml</univ-xml>
<univ-xml>univ2.xml</univ-xml>
<univ-xml>univ3.xml</univ-xml>
</univ-xml-list>
and the xsl should now go through each of the univ xmls and merge them.
What I thot of doing was first go through one of the xmls( the first one for 
example ) and construct the elements.

<xsl:for-each select="document(/univ-xml-list/univ-xml[1])">
<xsl:variable name = "element-name" select="name(.)"/>
<xsl:element name="$element-name">


</xsl:element>
</xsl:for-each>
this will create all the elements that are in the first elements..

now the next thing what i want to do is ..
run a for-each against the mainxml.xml and plug in the element data into the 
above created element "where" the element name equals the name of the 
element from the first for-loop

that is..
<xsl:for-each select="document(/univ-xml-list/univ-xml[1])">
<xsl:variable name = "element-name" select="name(.)"/>
<xsl:element name="$element-name">
<xsl:for-each select="document(/univ-xml-list/univ-xml)/university-records">
<xsl:copy-of select="*[name(.) = $element-name]"/>
</xsl:for-each>
</xsl:element>
</xsl:for-each>
my intention above was to fill in the data elements from all the xml files 
specified in the mainxml.xml into the element created using the first xml( 
all xmls will have same structure)

But this does not work..
one of the univ xmls is as  follows
<?xml version="1.0"?>
<university-records>
<univ-ids>
<univ id = "NSU">
<name> Newyork State University</name>
<location> Newyork </location>
</univ>
<univ id = "BU">
<name> Belmont University</name>
<location> Belmont </location>
</univ>
</univ-ids>
<university-results>
<university univ-id = "NSU">70%</university>
<university univ-id = "BU">60%</university>
</university-results>
</university-records>

What i noticed was.. When the for-each is processing one document(),
it is not able to process any other document.. and also fails to recognize 
the elements from the mainxml.

I do not know if there is any other way of doing this?
Thanks

> From: David Carlisle <davidc@nag.co.uk>
> Reply-To: xsl-list@lists.mulberrytech.com
> To: xsl-list@lists.mulberrytech.com
> Subject: Re: [xsl] document() loops
> Date: Thu, 12 Sep 2002 17:41:00 +0100
>
> You say you want to merge the files but your two loops only have one
> node each (the root nodes of the first and second documents respectively).
> So as written you don't really need a for-each at all as there is only
> one node being used.
>
> I posted some code before (I just noticed this is the same thread as the
> earlier question) does that not do what you expect?
>
> David
>
>
> _____________________________________________________________________
> This message has been checked for all known viruses by Star Internet
> delivered through the MessageLabs Virus Scanning Service. For further
> information visit http://www.star.net.uk/stats.asp or alternatively call
> Star Internet for details on the Virus Scanning Service.
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list



_________________________________________________________________
Join the world’s largest e-mail service with MSN Hotmail. 
http://www.hotmail.com


XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list