Question

I’m writing an XSLT template that need to output a valid xml file for an xml Sitemap.

<url>
<loc>
    <xsl:value-of select="umbraco.library:NiceUrl($node/@id)"/>
</loc>
<lastmod>
    <xsl:value-of select="concat($node/@updateDate,'+00:00')"/>
</lastmod>
</url>

Unfortunately, Url that is output contains an apostrophe – /what’s-new.aspx

I need to escape the ‘ to ' for google Sitemap. Unfortunately every attempt I’ve tried treats the string ‘'‘ as if it was ”’ which is invalid – frustrating. XSLT can drive me mad sometimes.

Any ideas for a technique? (Assume I can find my way around XSLT 1.0 templates and functions)

Answer 1

So you have ' in your input, but you need the string   in your output?

In your XSL file, replace ' with ', using this find/replace implementation (unless you are using XSLT 2.0):

<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>

Call it this way:

<loc>
  <xsl:call-template name="string-replace-all">
    <xsl:with-param name="text" select="umbraco.library:NiceUrl($node/@id)"/>
    <xsl:with-param name="replace" select="&apos;"/>
    <xsl:with-param name="by" select="&amp;apos;"/>
  </xsl:call-template>
</loc>

The problem is ' is interpreted by XSL as '. ' will be interpreted as '.