How to remove path

Olaf_l · April 11th, 2005, 06:53 AM

I have, for example, this string
'c:/programm/file/test.txt'

How do I remove the path of test.txt. All that should remain is 'test.txt'.

I tried the function substring-after like this

Code:

                    <afbeelding>
                        <xsl:variable name="bestand" select="image_reference"/>
                        <xsl:value-of select="substring-after($bestand,'/')"/>
                    </afbeelding>

but this removes everything after the first appearance of '/'. Is it possible to delete everything before the last '/' ??

barcher · April 11th, 2005, 07:03 AM

You can use a recursive template:

<xsl:template name="getFilename">
    <xsl:param name="filePath" />

    <xsl:variable name="rest-of" select="substring-after($filePath, '/')" />

    <xsl:choose>
        <xsl:when test="contains($rest-of, '/')">
            <xsl:call-template name="getFilename">
                <xsl:with-param name="filePath" select="$rest-of" />
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:value-of select="$rest-of" />
        </xsl:otherwise>
    </xsl:choose>

</xsl:template>

mhkay · April 11th, 2005, 07:13 AM

In XSLT 1.0, use a recursive template.

In XSLT 2.0, use

tokenize($filename, '/')[last()]

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Olaf_l · April 25th, 2005, 04:19 AM

I tried this code for tokenizing, but it doesn't work. Can someone take a look at this code??

Code:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; version="2.0">
....
<image>
   <xsl:value-of select="tokenize('image_reference, '/')[last()]" />
</image>

TIA

mhkay · April 25th, 2005, 11:15 AM

tokenize('
^

you haven't closed this opening quote.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

mhkay · April 25th, 2005, 11:16 AM

By the way, saying that something "doesn't work" isn't very helpful. Tell us what output you expected and what output you actually got.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Olaf_l · April 26th, 2005, 03:58 AM

My apologies

I use now this (partial) code

Code:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="/">
......

<image>
   <xsl:value-of select="tokenize('image_reference', '/')[last()]" />
</image>

The error that is given is:
'Tokenize is not a valid XSLT or XPATH function.
-->tokenize('image_reference', '/')<-- [last()]

An example of what I want is this:
The string:
'C:/Program Files/Adlib/data/afbeeldingen/0135.jpg'
Must become
'0135.jpg' or '/0135.jpg'

raman27 · April 26th, 2005, 04:06 AM

hi!

tokenize is a keyword available only in xslt version "2.0" But you are using the version="1.0", so change to version="2.0" and then try again
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

Regards
Seetharaman

Olaf_l · April 26th, 2005, 04:12 AM

The 1.0 is in the <xml> tag. The xsl-tag is already version 2.0.

Should i attach another namespace perhaps ?

raman27 · April 26th, 2005, 04:19 AM

hi!

Now Try this!

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:Str="http://">

<xsl:template match="/">
<xsl:variable name="t" select="'c:\\test\\out.txt'"/>
<xsl:value-of select="tokenize($t, '\\')[last()]"/>
</xsl:template>

</xsl:stylesheet>