- Author:
- limodou
- Posted:
- February 25, 2007
- Language:
- Python
- Version:
- Pre .96
- Score:
- 2 (after 2 ratings)
Convert plain text to html. For example:
text="""aabhttp://www.example.com http://www.example.com
http://www.example.com <<< [<aaaa></aaaa>] """ print plaintext2html(text)
It can convert url text to html href. And it also can convert space to . So if you paste python code, the indent will not lost at all. Default it'll convert t to 4 spaces.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 | import re
import cgi
re_string = re.compile(r'(?P<htmlchars>[<&>])|(?P<space>^[ \t]+)|(?P<lineend>\r\n|\r|\n)|(?P<protocal>(^|\s)((http|ftp)://.*?))(\s|$)', re.S|re.M|re.I)
def plaintext2html(text, tabstop=4):
def do_sub(m):
c = m.groupdict()
if c['htmlchars']:
return cgi.escape(c['htmlchars'])
if c['lineend']:
return '<br>'
elif c['space']:
t = m.group().replace('\t', ' '*tabstop)
t = t.replace(' ', ' ')
return t
elif c['space'] == '\t':
return ' '*tabstop;
else:
url = m.group('protocal')
if url.startswith(' '):
prefix = ' '
url = url[1:]
else:
prefix = ''
last = m.groups()[-1]
if last in ['\n', '\r', '\r\n']:
last = '<br>'
return '%s<a href="%s">%s</a>%s' % (prefix, url, url, last)
return re.sub(re_string, do_sub, text)
|
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Comments
you have a semicolon at the end of line 17. Seems like a syntax error
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