Subject: [xsl] whitespace as a parameter to a template (replace linefeed with li nefeed tab) From: "MacEwan, James (Information Services)" <James.MacEwan@xxxxxxxxxxxxxxxxxx> Date: Wed, 30 May 2001 11:58:26 -0500 |
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@xxxxxxxxxxxxxxxxxx v: (204) 956-8515 f: (204) 943-3540 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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