I have URLs with some special characters and i would like to replace them and I am using xslt 1.0 so I am writing the code as below.
The provided "replace" template is working.
To confirm this use the following transformation:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/*">
<xsl:variable name="link">
<xsl:call-template name="string-replace-all">
<xsl:with-param name="text" select="@FileRef"/>
<xsl:with-param name="replace">'</xsl:with-param>
<xsl:with-param name="by" select="'%27'"/>
</xsl:call-template>
</xsl:variable>
"<xsl:value-of select="$link"/>"
</xsl:template>
<xsl:template name="string-replace-all">
<xsl:param name="text"/>
<xsl:param name="replace"/>
<xsl:param name="by"/>
<xsl:choose>
<xsl:when test="contains($text,$replace)">
<xsl:value-of select="substring-before($text,$replace)"/>
<xsl:value-of select="$by"/>
<xsl:call-template name="string-replace-all">
<xsl:with-param name="text" select="substring-after($text,$replace)"/>
<xsl:with-param name="replace" select="$replace"/>
<xsl:with-param name="by" select="$by"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$text"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
when the transformation is applied on this XML document:
<t FileRef="Abc'Xyz'Tuv"/>
the wanted, correct result is produced:
"Abc%27Xyz%27Tuv"
The problem may be in the way you specify the replace
parameter (uneven number of apostrophes in the expression):
Instead of:
<xsl:with-param name="replace" select="'''"/>
Use:
<xsl:with-param name="replace">'</xsl:with-param>