Re: [xsl] document() loops

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

Sorry if i am dragging it a bit too much,
but,
<xsl:for-each 
select="document(univ-xml[1])/university-records/univ-ids/univ">
<xsl:value-of select="position()"/>
<br/>
<xsl:value-of select="name"/>
<hr/>
</xsl:for-each>
works fine .. iterates through the nodes..and gets the data from the xml 
file at position1 ie., univ-xml1.. perfect!!
but when i try to iterate through other XML files WITHIN the above for-each, 
nothin shows up..
example.
<xsl:template match="univ-xml-list">
<first-xml>
<xsl:for-each 
select="document(univ-xml[1])/university-records/univ-ids/univ">
<xsl:element name= "univ{position()}">
<xsl:value-of select="name"/>
</xsl:element>
<xsl:if test= "position() = last()">
<second-xml>
<xsl:for-each 
select="document(univ-xml[2])/university-records/univ-ids/univ">
<!-- TO ITERATE THROUGH THE 2nd XML -->
<xsl:element name= "univ{position()}">
<xsl:value-of select="name"/>
</xsl:element>
</xsl:for-each>
</second-xml>
</xsl:if>
</xsl:for-each>
</first-xml>
</xsl:template>

I was expecting to get a result like..
<?xml version="1.0" encoding="UTF-16"?>
<first-xml>
	<univ1>Kansas State University</univ1>
	<univ2>Florida Atlantic University</univ2>
	<univ3> Mississipi State University</univ3>
	<univ4>Ohio State University</univ4>
	<second-xml>
		<univ1>Ohio State University</univ1>
		<univ2>Wisconsin state University</univ2>
		<univ3> Rensellar polytechnic Institute </univ3>
		<univ4> Mississipi State University</univ4>
	</second-xml>
</first-xml>

but what prints out is..
<first-xml>
	<univ1>Kansas State University</univ1>
	<univ2>Florida Atlantic University</univ2>
	<univ3> Mississipi State University</univ3>
	<univ4>Ohio State University</univ4>
	<second-xml></second-xml>
</first-xml>

am i wrong anywhere ? if yes then what has to be done to get the result i 
was expecting.. this might probably lead me towards what i want..

thanks
> From: David Carlisle <davidc@nag.co.uk>
> Reply-To: xsl-list@lists.mulberrytech.com
> To: xsl-list@lists.mulberrytech.com
> CC: xsl-list@lists.mulberrytech.com
> Subject: Re: [xsl] document() loops
> your terminology does not fit well with that of XSLT so I am having to
> guess what you mean. for-each iterates over the set of nodes supplied,
> these can come from any number of files: you just have to select the
> appropriate nodes.
> Perhaps you would be better to apply-templates rather than do a for->each
> if you want to traverse into the file structure.
> But otherwise, as I think i said before, something like
> <xsl:for-each select="univ-xml">
>stuff from <xsl:value-of select="."/>
 ><xsl:for-each select="document(.)/university-records/*">
>  iterate over all the children of university-records in this file...




> From: "Laura Jenkins" <xsl_list@hotmail.com>
> Reply-To: xsl-list@lists.mulberrytech.com
> To: xsl-list@lists.mulberrytech.com
> Subject: Re: [xsl] document() loops
> Date: Fri, 13 Sep 2002 11:21:43 +0000
>
> oh ic what u mean.. then in such a case , how do we run a for loop for an 
> imported xml file???
> for.example
> if i get the locations from running a for-each to mainxml.xml
> and get hold of the first univ-xml's name..
> <xsl:for-each select="document(/univ-xml-list/univ-xml)">
> <xsl:variable name = "external-xml" select ="document(.)"/>
> </xsl:for-each>
>
> Now is there a way to run a for-each for this external xml file??
> so that i get hold of EACH ELEMENT in the External XML?? and get hold of 
> their properties like name() position() etc??
>
>
> > From: David Carlisle <davidc@nag.co.uk>
> > Reply-To: xsl-list@lists.mulberrytech.com
> > To: xsl-list@lists.mulberrytech.com
> > CC: xsl-list@lists.mulberrytech.com
> > Subject: Re: [xsl] document() loops
> > Date: Fri, 13 Sep 2002 10:19:21 +0100
> >
> >   <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..
> >
> > No it won't as I said before that for-each is iterating over a single node
> > so it will only create one element, with name equal to teh name of teh
> > node selected in the xsl:for-each select=" however that node is a
> > document node, so has name "" so the xsl:eleemnt should generate an
> > error thatyou are trying to generate an element with an invalid name,
> > "".
> >
> >
> > David
> >
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