javascript XML processing by XSL file not working

Question

In javascript, trying to transform a dynamically created XML data island, sorting it using a XSL file, but the result is the sorted data all on one line, without XML formatting or proper indenting. It looks like the is not being used at all. I need the XML tags and indenting to be generated in the resulting transformNode().

javascript code:

var sourceXML = document.getElementById(XMLViewID); //textArea containing XML
var xmlDoc=new ActiveXObject("Microsoft.XMLDOM");
xmlDoc.async = false;
xmlDoc.loadXML(sourceXML.value);

var xslDoc = new ActiveXObject("Microsoft.XMLDOM");
xslDoc.async=false;
xslDoc.load("xsl.xsl");

// This should be the sorted, formatted XML data, in tree and indented format?
var sorted = xmlDoc.transformNode(xslDoc);

XML DATA:

    <table>
    <row>
        <A>j</A>
        <B>0</B>
    </row>
    <row>
        <A>c</A>
        <B>4</B>
    </row>
    <row>
        <A>f</A>
        <B>6</B>
    </row>
</table>

xsl.xsl:

<xsl:output method="xml" indent="yes" omit-xml-declaration="no"/>

<xsl:template match="/">
    <xsl:apply-templates select="table/row">
        <xsl:sort select="A" order="ascending"/>
    </xsl:apply-templates>
</xsl:template>

<xsl:template match="row">
    <xsl:value-of select="A"/>
    <xsl:value-of select="B"/>
</xsl:template>

I assume with 'indent=yes' and 'omit-xml-declaration=no' that the resulting transformation should be with indenting and formatting:

 <?xml version="1.0" encoding="UTF-16"?>
     <table>
       <row>
         <tr>
           <A>j</A>
           <B>0</B>
         </tr>
         <tr>
           <A>c</A>
           <B>4</B>
         </tr>
         <tr>
           <A>f</A>
           <B>6</B>
         </tr>
       </row>
     </table>

But instead it is: c4f6j0 in one line, no formatting, no XML tags...

Martin Honnen · Accepted Answer · 2013-08-28 09:03:54Z

1

So far your XSLT produces nothing but text nodes, if you want XML with elements then you need to create them with code like

<xsl:template match="table">
  <xsl:copy>
    <xsl:apply-templates select="row">
      <xsl:sort select="A"/>
    </xsl:apply-templates>
  </xsl:copy>
</xsl:template>

<xsl:template match="@* | node()">
  <xsl:copy>
    <xsl:apply-templates select="@* | node()"/>
  </xsl:copy>
</xsl:template>

answered Aug 28, 2013 at 9:03

Martin Honnen

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3 Comments

z-- Over a year ago

The template with match="@* | node() is the identity transform. It makes sure that nodes and attributes not mentioned in other templates are copied as is. more

Andy Over a year ago

That worked great to get the xml tags in there. I think I already know the answer to my next question, but is there a way for the indented formatting to be put into the sorted string as well? The result I get back is everything expected but without the indentation. My guess is that is up to whatever is displaying the resulting xml, but it would be great if there were a way for the xslt to add that automatically.

z-- Over a year ago

Try to figure it out by looking at the answers of stackoverflow.com/search?q=[xslt]+white+space+preserve .

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