Subject: [xsl] Access XML from the URL and put in different file From: Shashank Jain <shashankjain@xxxxxxxx> Date: Wed, 13 Apr 2011 12:15:39 -0500 |
Hello All, Following is the test xml I am working on <Description> <core resource="http://xyz.com/93034B4E-5686-4FF5-9EAD-F603F42B897F/viewas/xml"/> <core resource="http://xyz.com/E4E0EF48-030F-4ACE-B15E-E923AC582A6B/viewas/xml"/> <core resource="http://xyz.com/9808A0E2-5678-435F-8C89-5B7C8C27E380/viewas/xml"/> <core resource="http://xyz.com/80E137ED-E8CC-4B14-AC47-E437D8E91BBA/viewas/xml"/> <core resource="http://xyz.com/97337AAC-81CF-4DD8-9C8F-B06744B13753/viewas/xml"/> <core resource="http://xyz.com/114F7C46-C207-49BE-96D1-B5E4314342B9/viewas/xml"/><D escription> The value of resource attribute is a link to the XML document. I need to write an XSLT which creates 6 different output file (as there are 6 <core>), and every output file should have their respective XML. =========This is the test.xslt I wrote<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions"> <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" name="xml"/> <xsl:template match="/"> <xsl:for-each select="//core"> <xsl:variable name="filename" select="concat('output1/',substring(substring-after(@*:resource,'xyz.com/'),1 ,5),'.xml')"/> <xsl:value-of select="$filename"/> <!-- Creating --> <xsl:result-document href="{$filename}" format="xml"> <xsl:value-of select="@*:resource"/> </xsl:result-document> </xsl:for-each> </xsl:template></xsl:stylesheet>===========I was able to get 6 different xml files, but they were having the value of the respective resource attribute, not the whole XML. Output========<?xml version="1.0" encoding="UTF-8"?>http://xyz.com/93034B4E-5686-4FF5-9EAD-F603F42B897F/viewas/ xml Required output is to get the whole XML, which appears when the link is clicked/navigated. Is there any way I can achieve this using XSLT. Thanks,Shashank Jain
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