RE: [xsl] Finding the path from the filename

Subject: RE: [xsl] Finding the path from the filename
From: Cams Ismael <Ismael.Cams@xxxxxxxxxxxxxxx>
Date: Mon, 20 Jan 2003 16:35:01 +0100
What you need is the position of the last \ character. So first write a
template that returns you the position of the last \. Therefore you need
something like this:

<xsl:template name="lastCharPosition">
	<xsl:param name="original"/>
	<xsl:param name="character"/>
	<xsl:variable name="char_len">
		<xsl:value-of select="string-length($character)"/>
	</xsl:variable>
	<xsl:choose>
		<xsl:when test="contains($original,$character)">
			<xsl:choose>
				<xsl:when
test="contains(substring-after($original,$character),$character)">
					<xsl:call-template
name="lastCharPosition">
						<xsl:with-param
name="original" select="substring-after($original,$character)"/>
						<xsl:with-param
name="character" select="$character"/>
					</xsl:call-template>
				</xsl:when>
				<xsl:otherwise>
					<xsl:value-of
select="string-length(substring-before($original,$character))+$char_len"/>
				</xsl:otherwise>
			</xsl:choose>
		</xsl:when>
		<xsl:otherwise>
			<xsl:value-of select="string-length($original)"/>
		</xsl:otherwise>
	</xsl:choose>
</xsl:template>

If you have this position you can use the substring-before() function to
return everything before the file name.

Hope this helps.

Kind regards,
Ismaël

-----Original Message-----
From: Thomas V. Nielsen [mailto:thomas@xxxxxxxxxxxxx]
Sent: maandag 20 januari 2003 16:18
To: XSL List
Subject: [xsl] 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

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