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POUse XSLT to pretty print XML/XHTML without corrupting namespace info
 

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  1. POUse XSLT to pretty print XML/XHTML without corrupting namespace info
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    <p>I'm trying to use XSLTs (in Javascript) to pretty-print an XHTML doc that has been machine generated. However, the various XSLTs I've tried to use, all mangle the xmlns attributes (see below).</p> <p>Here is a sample of desired output (made by hand from unindented, compact, XHTML). </p> <pre><code>&lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;h:html xmlns:h="http://www.w3.org/1999/xhtml" xmlns:orx="http://openrosa.org/jr/xforms" xmlns="http://www.w3.org/2002/xforms" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:jr="http://openrosa.org/javarosa"&gt; &lt;h:head&gt; &lt;h:title&gt;New Form1&lt;/h:title&gt; .... </code></pre> <p>Here is what I'm getting instead:</p> <pre><code>&lt;h:html h="http://www.w3.org/1999/xhtml" orx="http://openrosa.org/jr/xforms" xmlns="http://www.w3.org/2002/xforms" xsd="http://www.w3.org/2001/XMLSchema" jr="http://openrosa.org/javarosa"&gt; &lt;h:head&gt; &lt;h:title&gt;New Form1&lt;/h:title&gt; ... </code></pre> <p>Notice the xmlns attributes are altered in the 'h:html' tag in the second code snippet. Also the beginning <code>&lt;?xml ...&gt;</code> tag is missing.</p> <p>This is (one of many) XSLTs I've used with similar results:</p> <pre><code>&lt;xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;\ &lt;xsl:output method="xml" indent="yes"/&gt; &lt;xsl:strip-space elements="*"/&gt; &lt;xsl:template match="/"&gt; &lt;xsl:copy-of select="."/&gt; &lt;/xsl:template&gt; &lt;/xsl:stylesheet&gt; </code></pre> <p>Any ideas on what I'm doing wrong? Am I trying to do the impossible?</p> <p>If you're wondering why I'm trying to do this: I have to use GWT as the framework for designing a FormDesigner web app. This is the output, but needs to be human readable for the more technically inclined users that want to do by-hand edits. GWT just <em>doesn't do</em> xml pretty printing (as far as I can tell in my searching so far). Thus, we go native to JS land and try for a solution there.</p> <p>Ideas/solutions would be greatly appreciated!</p> <p>Edit:</p> <p>Here is the Javascript that makes use of the XSLT. I call the beautifyXML() function to actual perform the indentation:</p> <pre><code>//var xsl_string = '&lt;xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt; &lt;xsl:output omit-xml-declaration="yes" indent="yes"/&gt; &lt;xsl:template match="node()|@*"&gt; &lt;xsl:copy&gt; &lt;xsl:apply-templates select="node()|@*"/&gt; &lt;/xsl:copy&gt; &lt;/xsl:template&gt;&lt;/xsl:stylesheet&gt;'; // from: http://www.xml.com/pub/a/2006/11/29/xslt-xml-pretty-printer.html?page=3 var xsl_string = '&lt;xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;\ &lt;xsl:output method="xml" indent="yes"/&gt;\ &lt;xsl:strip-space elements="*"/&gt;\ &lt;xsl:template match="/"&gt;\ &lt;xsl:copy-of select="."/&gt;\ &lt;/xsl:template&gt;\ &lt;/xsl:stylesheet&gt;'; var xsl = (new DOMParser()).parseFromString(xsl_string, "text/xml"); function stringToXml(xml_string) { return (new DOMParser()).parseFromString(xml_string, "text/xml"); } function xmlToString(xml) { return (new XMLSerializer()).serializeToString(xml); } function isParseError(xml) { try { // console.log( xml.documentElement.firstChild.firstChild.tagName); return xml.documentElement.tagName == "parsererror" || xml.documentElement.firstChild.firstChild.tagName == "parsererror"; } catch (ex) { return false; } } function beautifyXml(input) { var xml = stringToXml(input); if (isParseError(xml)) { return input; } var transformedXml = xslTransformation(xml, xsl); return xmlToString(transformedXml); } /** * @param xml * @param xsl */ function xslTransformation(xml, xsl) { // code for IE if (window.ActiveXObject) { var ex = xml.transformNode(xsl); return ex; } // code for Mozilla, Firefox, Opera, etc. else if (document.implementation &amp;&amp; document.implementation.createDocument) { var xsltProcessor = new XSLTProcessor(); xsltProcessor.importStylesheet(xsl); var resultDocument = xsltProcessor.transformToFragment(xml, document); return resultDocument; } } </code></pre>
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    1. COIs not corrupt an overstatement? The XML is valid?
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    2. COHow are you applying the transform? Can you post the JavaScript, or is it buried in auto-generated GWT code? Your XSLT is fine(other than the trailing "\" inside of the `xsl:stylesheet` element.
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    3. CO@Thomas, the xml is valid, but doesn't make sense. xmlns:orx != orx in terms of what they mean as attributes.
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