Subject: [xsl] replacement in more than one Tag in node From: henry human <henry_human@xxxxxxxx> Date: Wed, 30 Mar 2005 00:00:53 +0200 (CEST) |
hello, this time i will to define a template to replace . with : in more than one Tags f.e replace in name, surname,aaa,bbb ... how should the template rule be defined to do this?? <?xml version="1.0"?> <?xml-stylesheet type="text/xsl" href="my.xsl"?> <document> <part1> <ttt>u.bb</ttt> <name>u.bb</name> <surname>u.bb</surname> <aaa>u.bb</aaa> <bbb>u.bb</bbb> </part1> <part2> <name>u.bb</name> </part2> </document> ----- <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html"/> <xsl:template match="document"> <xsl:apply-templates select="part1"/> </xsl:template> <xsl:template match="ptr1"> <xsl:call-template name="part1"> <xsl:with-param name="name" select="name"/> <xsl:with-param name="search_for" select="'.'"/> <xsl:with-param name="replace_out" select="':'"/> </xsl:call-template> </xsl:template> <xsl:template name="part1"> <xsl:param name="name"/> <xsl:param name="search_for"/> <xsl:param name="replace_out"/> <xsl:choose> <xsl:when test="contains($name,$search_for)"> <xsl:value-of select="substring-before($name,$search_for)"/> <xsl:value-of select="$replace_out"/> <xsl:call-template name="part1"> <xsl:with-param name="name" select="substring-after($name,$search_for)"/> <xsl:with-param name="search_for" select="$search_for"/> <xsl:with-param name="replace_out" select="$replace_out"/> </xsl:call-template> </xsl:when> <xsl:otherwise> <xsl:value-of select="$name"/> </xsl:otherwise></xsl:choose></xsl:template> </xsl:stylesheet> ___________________________________________________________ Gesendet von Yahoo! Mail - Jetzt mit 250MB Speicher kostenlos - Hier anmelden: http://mail.yahoo.de
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