In 2.0:
select="document(document('index.xml')/protest/package/name/concat(.,
'.xml'))">
In 1.0:
<xsl:for-each select="document('index.xml')/protest/package/name">
<xsl:for-each select="document(concat(., '.xml'))">
....
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: chun ji [mailto:cji_work(_at_)yahoo(_dot_)com]
Sent: 04 December 2006 23:27
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] new bie question
I have an index.xml file, which reads as:
"
-<?xml version="1.0" encoding="ISO-8859-1" ?>
-<?xml-stylesheet href="rules.xsl" type="text/xsl"?> -<protest>
-<package>
-<name>billing</name>
-</package>
-<package>
-<name>timecard</name>
-</package>
-</protest>
".
There is a group of child xml files that associated with each
"/protest/package/name", such as "billing.xml", "timecard.xml".
Now I have soming in my XSL file that tries to open all these
CHILD xml files, " <xsl:for-each
select="document(document('index.xml')/protest/package/name)">
.
</xsl:for-each>
"
which fails, because I did not give ".xml" for each
child xml.
Does someone know the format to do it ?
thanks a lot
cji
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