Cas, search this list for references to Sal Mangano's "XSL Cookbook"; it
contains templates for text-replace, split(), etc.
--A
From: Cas Tuyn <Cas(_dot_)Tuyn(_at_)asml(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] translate array delimiters in XSLT
Date: Fri, 18 Mar 2005 17:36:55 +0100
Hi,
To display a list of software releases with ", " delimited that are
mentioned in XML as ";" delimited:
<procedure software="3.3.7;3.3.8;3.3.9;3.4.0;3.4.2;3.4.4">
I had written in the perl function that cleans up the XML, this sinple
regular expression:
if ( $line =~ /software=/ ) {
$line =~ s#;#, #g ;
}
Obviously this only worked for software values on the same line and not
for software values that spanned multiple lines, so I now want to
transfer this 'logic' from Perl to XSLT. I removed the 3 lines from the
Perl program, and changed this XSLT:
<xsl:for-each select="@software">
<xsl:value-of select="."/>
</xsl:for-each>
to:
<xsl:for-each select="@software">
<xsl:value-of select="."/>,
</xsl:for-each>
but that only produced one "," behind the last SW release, not inbetween.
Reading in Michael Kay's XSLT book on the translate() function made me
try this:
<xsl:for-each select="translate(@software,';',', ')">
<xsl:value-of select="."/>
</xsl:for-each>
but that did not prove valid XSLT, even with string() around the array
@software.
Anyone has a tip to convert ";" to ", " without reverting to Perl?
Cas
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