Re: [xsl] xslt replace special characters

Subject: Re: [xsl] xslt replace special characters
From: Greg Faron <gfaron@xxxxxxxxxxxxxxxxxx>
Date: Fri, 08 Nov 2002 12:51:21 -0700
At 12:40 PM 11/8/2002, you wrote:
How do you replace characters in xslt.

i have a url string that may contain '?' , '&', and '='
i want to convert

? to %3f
= to %3d
& to %26

is there a good way to do this? i try using translate function but it doesn't work that well. i've also tried writing a recursive template that checks for all occurence of these characters and it doesnt seem to work.
any idea?

<xsl:template name="replace">
<xsl:param name="string" select="''"/>
<xsl:param name="pattern" select="''"/>
<xsl:param name="replacement" select="''"/>
<xsl:when test="$pattern != '' and $string != '' and contains($string, $pattern)">
<xsl:value-of select="substring-before($string, $pattern)"/>
Use "xsl:copy-of" instead of "xsl:value-of" so that users
may substitute nodes as well as strings for $replacement.
<xsl:copy-of select="$replacement"/>
<xsl:call-template name="replace">
<xsl:with-param name="string" select="substring-after($string, $pattern)"/>
<xsl:with-param name="pattern" select="$pattern"/>
<xsl:with-param name="replacement" select="$replacement"/>
<xsl:value-of select="$string"/>

Greg Faron Integre Technical Publishing Co.

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