RE: [xsl] Result tree fragment to string?

From: David Carlisle [mailto:davidc(_at_)nag(_dot_)co(_dot_)uk]
Sent: Wednesday, August 27, 2008 12:20 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Result tree fragment to string?

Do I have any options left for converting the result tree fragment to

string?

no


Hmm... could he do something like:

<xsl:variable name="xml">
  <xsl:call-template name="serialize-tree">
    <xsl:with-param name="nodes" select="$result-tree" />
  </xsl:call-template>
</xsl:variable>

<xsl:template name="serialize-tree">
  <xsl:param name="nodes" />
  <xsl:apply-templates select="$nodes" mode="serialize" />
</xsl:template>

<xsl:template name="quote-attribute">
  <xsl:param name="value" />
  <!-- quote attribute's value -->
</xsl:template>

<xsl:template match="node()" mode="serialize" />

<xsl:template match="@*" mode="serialize">
  <xsl:variable name="dq">
    <xsl:text>"</xsl:text>
  </xsl:variable>
  <xsl:variable name="value">
    <xsl:call-template name="quote-attribute">
      <xsl:with-param name="value" select="." />
    </xsl:call-template>
  </xsl:variable>
  <xsl:value-of select="concat('&#32;',local-name(),'=',$dq,$value,$dq)" />
</xsl:template>

<xsl:template match="*" mode="serialize">
  <xsl:value-of select="concat('&lt;',local-name())" />
  <xsl:apply-templates select="@*" mode="serialize" />
  <xsl:value-of select="string('&gt;')" />
  <xsl:apply-templates select="*" mode="serialize" />
  <xsl:value-of select="concat('&lt;/',local-name(),'&gt;')" />
</xsl:template>

The above doesn't handle namespaces, processing-instructions, comments,
or mixed content, but it could be hacked to do so.  Is there a reason
why this approach would not work in XSL 1.0 to satisfy his needs?


Andy.



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