XSLT Check if directory exist

ArnArn · #1 (**permalink**) November 21st, 2008, 03:09 PM

Hi,

Is there a way to check if a certain directory in unix exists using xslt?

Sample:

//phome/folder/subolder1
//phome/folder/subolder2
//phome/folder/subolder3
//phome/folder/

If found this in p2p and tested it but it doesnt seem to work:

(image.xsl)

<xsl:template name="xml-file-exists">
    <xsl:param name="url"/>
    <xsl:variable name="test-url">//(unix_diretory)<xsl:value-of select="$url"/>
    </xsl:variable>
    <xsl:copy-of select="document($test-url)//file"/>
  </xsl:template>

  <xsl:variable name="file-exists">
    <xsl:call-template name="xml-file-exists">
      <xsl:with-param name="url" select="$options.xml"/>
    </xsl:call-template>
  </xsl:variable>

  <xsl:choose>
     <xsl:when test="$file-exists='true'">
     ...xsl code using document()...
     </xsl:when>
     <xsl:otherwise>
     ...xsl code without document()
     </xsl:otherwise>
  </xsl:choose>

I need to check if the folder exist in unix and if it is it will be uploaded in a certain site using ant:

<ftsfileupload verbose="no" failOnError="no" ftsURL="${ftsURL}" stylesheet="${images}" rootName="root" generateIdx="true" >
    <fileset dir="${graphicsDir}" includes="**/*"/>
</ftsfileupload>

thanks...

Martin Honnen · #2 (**permalink**) November 22nd, 2008, 08:23 AM

I don't think pure XSLT can check for the existance of a file or folder. You need an extension function or extension object for that. How you do that exactly is then highly processor dependent.

--
Martin Honnen
Microsoft MVP - XML

Martin Honnen · #3 (**permalink**) November 22nd, 2008, 08:43 AM

Just to be clear, if you want to check whether an XML document can be loaded, then XSLT 2.0 has http://www.w3.org/TR/xpath-functions...-doc-available. But I don't think that helps to check for the existance of a folder/directory.

--
Martin Honnen
Microsoft MVP - XML

bonekrusher · #4 (**permalink**) November 23rd, 2008, 03:44 PM

Hi,

If you need to check if a file exist, use an extension. What type of extension you use depends on the xslt processor. In the below example, I use Saxon 9.4, which use java.

Create and compile the java class:

Code:

import java.io.File;

public class FileExist {

    private static boolean exist;

    public static boolean fileExist(String myfile) {

        exist = new File(myfile).exists();

        return exist;

    }
}

XML:

Code:

<root>
    <member>2008-01-01</member>
    <member>2009-06-23</member>
    <file>C:\\temp2\\chap1.xml</file>
</root>

xslt:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ACM="java:FileExist">
    <xsl:output method="html"/>
    <xsl:template match="/">
        <html>
            <head>                
            </head>
            <body>
                <xsl:for-each select="root/file">
                    <h1>
                        <xsl:value-of select="ACM:fileExist(.)"/>
                    </h1>
                </xsl:for-each>
            </body>
        </html>
    </xsl:template>
</xsl:stylesheet>

Windows Command line to run:

Code:

@echo off
:: Run with extensions
:: place all classes in classes folder in your saxon direcory
java -classpath ./classes;saxon9.jar net.sf.saxon.Transform -s:"C:\input.xml" -xsl:"C:\java_ext_file_exist.xslt" -o:output.html & pause

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