68 lines
1.9 KiB
Python
68 lines
1.9 KiB
Python
# -*- coding: utf-8 -*-
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"""
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The Pygments Markdown Preprocessor
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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This fragment is a Markdown_ preprocessor that renders source code
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to HTML via Pygments. To use it, invoke Markdown like so::
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from markdown import Markdown
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md = Markdown()
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md.textPreprocessors.insert(0, CodeBlockPreprocessor())
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html = md.convert(someText)
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markdown is then a callable that can be passed to the context of
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a template and used in that template, for example.
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This uses CSS classes by default, so use
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``pygmentize -S <some style> -f html > pygments.css``
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to create a stylesheet to be added to the website.
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You can then highlight source code in your markdown markup::
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[sourcecode:lexer]
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some code
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[/sourcecode]
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.. _Markdown: http://www.freewisdom.org/projects/python-markdown/
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:copyright: Copyright 2006-2013 by the Pygments team, see AUTHORS.
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:license: BSD, see LICENSE for details.
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"""
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# Options
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# ~~~~~~~
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# Set to True if you want inline CSS styles instead of classes
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INLINESTYLES = False
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import re
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from markdown import TextPreprocessor
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from pygments import highlight
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from pygments.formatters import HtmlFormatter
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from pygments.lexers import get_lexer_by_name, TextLexer
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class CodeBlockPreprocessor(TextPreprocessor):
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pattern = re.compile(
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r'\[sourcecode:(.+?)\](.+?)\[/sourcecode\]', re.S)
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formatter = HtmlFormatter(noclasses=INLINESTYLES)
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def run(self, lines):
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def repl(m):
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try:
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lexer = get_lexer_by_name(m.group(1))
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except ValueError:
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lexer = TextLexer()
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code = highlight(m.group(2), lexer, self.formatter)
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code = code.replace('\n\n', '\n \n').replace('\n', '<br />')
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return '\n\n<div class="code">%s</div>\n\n' % code
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return self.pattern.sub(
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repl, lines)
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