Subject: Re: [xsl] Sorting XML data by <entry ID=""> From: "Charles Muller acmuller@xxxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Wed, 5 Jan 2022 04:51:37 -0000 |
The below XSLT seems to work, but despite the xsl:output method="xml" declaration, the files come out as .html (even though the contents are fine).
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
B <xsl:output method="xml" omit-xml-declaration="yes" indent="no"/>
<xsl:template match="dealt">
B <xsl:copy>
B <xsl:apply-templates select="@*"/>
B <xsl:apply-templates select="entry">
B B <xsl:sort select="@ID" order="ascending"/>
B </xsl:apply-templates>
B </xsl:copy>
</xsl:template>
<xsl:template match="@* | node()"> B <xsl:copy> B <xsl:apply-templates select="@* | node()"/> B </xsl:copy> </xsl:template>
Regards,
Chuck
--------------------------- A. Charles Muller, Professor
Buddhist Culture Research Center Musashino University 1-1-20 Shinmachi, Nishitokyo-shi Tokyo 202-8585, Japan Office Phone: 042-468-3145
(University of Tokyo, Professor Emeritus) Web Site: Resources for East Asian Language and Thought http://www.acmuller.net
Twitter: @ACharlesMuller
--------------------------- A. Charles Muller, Professor
Buddhist Culture Research Center Musashino University 1-1-20 Shinmachi, Nishitokyo-shi Tokyo 202-8585, Japan Office Phone: 042-468-3145
(University of Tokyo, Professor Emeritus) Web Site: Resources for East Asian Language and Thought http://www.acmuller.net
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