[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Finding the path from the filename
Hi Thomas. Try this: <xsl:template name="FullPath"> <xsl:param name="PathString" select="''"/> <xsl:if test="contains($PathString,'\')"> <xsl:value-of select="concat(substring-before($PathString,'\'),'\')"/> <xsl:call-template name="FullPath"> <xsl:with-param name="PathString" select="substring-after($PathString,'\')"/> </xsl:call-template> </xsl:if> </xsl:template> To call it just do: <xsl:call-template name="FullPath"> <xsl:with-param name="PathString" select="$local_path"/> </xsl:call-template> Where $local_path is a variable or a param with your path string -----Original Message----- From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of Thomas V. Nielsen Sent: segunda-feira, 20 de Janeiro de 2003 15:18 To: XSL List Subject: Finding the path from the filename In a XSLT I receive a parameter with a full path and file name, like; C:\Data\Test\File.xml I have tried fumbling with substring and substring-before, but with no luck. What I need is the full path without the file name, like C:\Data\Test\ Sometimes parameter looks like this ..\Test\File.xml And also here I need to find the path, like ..\Test\ Any suggestions in how to use the substring with some iteration? /Thomas XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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