Subject: Re: [xsl] bat file creation based on xsl /xml From: "Michele R Combs mrrothen@xxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Mon, 14 Jan 2019 18:00:05 -0000 |
If I read this right, it looks like you just want to generate a list of the XML files in a directory, is that correct? If youbre in a windows enviroment the bdirb command will do what you want. Michele From: Rahul Singh rahulsinghindia15@xxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Sent: Monday, January 14, 2019 11:45 AM To: XSL-List: The Open Forum on XSL <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>; xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] bat file creation based on xsl /xml Hi, I want to create .bat file for below logic, i have xml input in 1 directory. how to do?: <?xml version="1.0" encoding="UTF-8"?> <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" exclude-result-prefixes="xs fn"> <xsl:output method="text" version="1.0" encoding="UTF-8" indent="no"/> <xsl:template match="/"> <xsl:for-each select="collection('.?select=*.xml')"> <xsl:text> </xsl:text> <xsl:value-of select="document-uri(.)"/> </xsl:variable> </xsl:for-each> </xsl:template> </xsl:stylesheet> XSL-List info and archive<http://www.mulberrytech.com/xsl/xsl-list> EasyUnsubscribe<http://lists.mulberrytech.com/unsub/xsl-list/1127818> (by email<>)
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