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whitespace as a parameter to a template (replace linefeed with linefeed tab)
- To: "'xsl-list at lists dot mulberrytech dot com'" <xsl-list at lists dot mulberrytech dot com>
- Subject: [xsl] whitespace as a parameter to a template (replace linefeed with linefeed tab)
- From: "MacEwan, James (Information Services)" <James dot MacEwan at investorsgroup dot com>
- Date: Wed, 30 May 2001 11:58:26 -0500
- Reply-To: xsl-list at lists dot mulberrytech dot com
Hi xsl-list,
Background:
I am producing a text document and need to replace a linefeed character with
a linefeed tab combination.
My input XML "street" tag looks something like this:
...
<street><![CDATA[Suite 123
456 Harrington Way]]></street>
...
I want my output to look like this
...
Addr: Suite123
456 Harrington Way
...
I have roughly used the Replace string algorithm as found in Mike Kay's book
(see also http://www.dpawson.co.uk/xsl/replace.html). I do not wish to
hardcode in the values of the substring to find and substring to put in its
place, but to pass them in as parameters to a template.
Problem:
I am making the call to my "do-replace" template as follows:
<xsl:call-template name="do-replace">
<xsl:with-param name="text" select="$x"/>
<xsl:with-param name="replace"> </xsl:with-param>
<xsl:with-param name="by"> 	</xsl:with-param>
</xsl:call-template>
I suspect that the XML parser is converting my the whitespace in the
"replace" and "by" parameters to a single space or an empty string. What is
the proper way to preserve the white space in parameters being passed into a
template?
Thanks,
J.
James MacEwan
Software Developer
Investors Group Inc.
mailto:James.MacEwan@investorsgroup.com
v: (204) 956-8515
f: (204) 943-3540
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