XML Source View
Syntax highlighting for XML files with XSLT
This stylesheet package contains XSLT styles for syntax highlighting of
arbitrary XML files.
P a r a m e t e r s :
=====================
* format: controls, whether the output should be pretty-printed or tried to be
kept as near as possible to the original source. Default is to apply
formatting.
* base-indent: sets the indentation step for each level, if the output should
be formatted. The default are two spaces.
* style: The name of a stylesheet (without extension) to be used for display.
Note, that the content of the stylesheet, though CSS otherwise, must be
encircled by an element <css/> in the empty namespace.
H o w t o D e p l o y :
===========================
a) in Firefox: Add the following lines to your XML file:
<?xslt-param name="format" select="true()" ?>
<?xml-stylesheet type="text/xsl" href="view-source.xsl"?>
(other browsers don't support <?xslt-param ?>, you have to touch
view-source.xsl itself there.)
b) via a command line XSLT processor:
$ saxon -s:source.xml -xsl:view-source.xsl -o:out.xhtml
$ xalan -IN source.xml -XSL view-source.xsl -OUT out.xhtml
c) inside PHP:
<?php
$xsl = new DOMDocument;
$xsl->load('view-source.xsl');
$proc = new XSLTProcessor;
$proc->importStyleSheet($xsl);
$xml = new DOMDocument;
$xml->load('source.xml');
$proc->setParameter('', 'format', TRUE);
$proc->transformToURI($xml, 'file:///tmp/out.xhtml');
?>
d) in Python with libxml2 and libxslt bindings:
#! /usr/bin/env python
import libxml2, libxslt
styledoc = libxml2.parseFile("view-source.xsl")
style = libxslt.parseStylesheetDoc(styledoc)
doc = libxml2.parseFile('source.xml')
result = style.applyStylesheet(doc, {"format": True})
out = open('out.xhtml', 'w')
out.write(result.serialize())
style.freeStylesheet()
doc.freeDoc()
result.freeDoc()
out.close()
L i c e n s e :
===============
The stylesheet is published under an MIT-style license and the GPL v2. Choose
at your liking.