@Johnson1s
86 sats stacked
stacking since: #570165
bio
7 items
#!/usr/bin/python2.2 import re

from ruby's mombo:

def escape(x): return x.replace('&', '&').replace('<', '<').replace('>', '>')
def sanitize(body): body=escape(body)
# passthru <a href>, <em>, <i>, <b>, <blockquote>, <br/>, <p>
body=re.sub('&lt;a href="([^"]*)"&gt;([^&]*)&lt;/a&gt;',
	    '<a href="\\1">\\2</a>', body)
body=re.sub('&lt;a href=\'([^\']*)\'&gt;([^&]*)&lt;/a&gt;',
	    '<a href="\\1">\\2</a>', body)
return body
def handle(x): rd = (("``", '“'), ("''",'”'), ("`", '‘'), ("'", '\xe2\x80\x99'), ('---', '\xe2\x80\x93'), ('--', '\xe2\x80\x94'), (' \n',' '), ('\n','<br />\n')) # @@ these two are sorta backwards, but huffman coded... # @@ need some way to escape and ignore in <code>/<pre> for k,v in rd: x = x.replace(k, v)
x = re.sub(r'\b_(.*?)_\b', r'<em>\1</em>', x)

# @@ I have no idea why I have to use \B here and \b back there:
x = re.sub(r'\B\*(.*?)\*\B', r'<strong>\1</strong>', x)
x = re.sub(r'\B\|(.*?)\|\B', r'<code>\1</code>', x)

return x
def atx(x, full=1): #x = sanitize(x) if x and x[-1] == "\n": x = x[:-1] # trim closing \n if exists
paras = x.split('\n\n') #@@ requires all in memory
nextp, title = None, ''
for i in xrange(len(paras)): 
	p = paras[i]
	
	if p == '': continue # blank line

	elif p[0] == '$' or nextp == 'pre': # <pre>
		p = "<pre>"+p+"</pre>"
		nextp = None

	elif re.match(r'^\#+ ', p): # <h?>
		n=0
		while p[n] == '#': n+=1
		if n==1: title = p[n:].strip()
		p = "<h"+`n`+">"+handle(p[n:].strip())+"</h"+`n`+">"

	elif re.match(r'^ *(\*|\d+\.) ', p): # <ul>/<ol>
		#@@ should really do <li><p> for paragraphed lists
		if p.strip()[0] == '*': mode = 'ul'
		else: mode = 'ol'
		
		lines = p.split('\n')

		li = 0
		while li < len(lines):
			l = lines[li]
						
			if (mode == 'ul' and l[0] != '*' and l[:2] != ' *') or \
			   (mode == 'ol' and not re.match(r'^ *\d+\.', l)):
				del lines[li]
				lines[li-1] = lines[li-1] + l
			else:
				li += 1
				
		for li in xrange(len(lines)):
			l = lines[li].strip()

			if mode == 'ul' and l[0] == '*': l = l[1:]
			else: l = re.sub(r'^ *\d+\.', '', l)
			l = '  <li>'+handle(l.strip())+'</li>'

			lines[li] = l

		p = '<'+mode+'>\n'+'\n'.join(lines)+'\n</'+mode+'>'
		
	elif p[:3] == '   ': # <blockquote>
		p = "<blockquote>"+handle(p)+"</blockquote>"
		
	else: # <p>
		if p[-2:] == '::': nextp = "pre"; p = p[:-1]
		p = "<p>"+handle(p)+"</p>"
	
	paras[i] = p
	
doc = '\n\n'.join(paras)
if full: doc = """<html xmlns="http://www.w3.org/1999/xhtml"><head>
<title>"""+title+"""</title> <link rel="stylesheet" type="text/css" href="/style.css" /> </head><body>
"""+'\n\n'.join(paras) + """
</body></html>""" 
return doc
if name == "main": import sys if len(sys.argv) <= 1 or sys.argv[1] == "-": print atx(sys.stdin.read()), else: print atx(open(sys.argv[1]).read()),
""" TODO: smarter pants, generalized phrasals, prime characters (4'3", 80's) ""03cc1d0932bb99b0697f5b5e5961b83ab7fd66f1efc4c9f5c7bad66c1bcbe78f02@52.5.194.83:9735
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0 sats \ 0 replies \ @Hamstr 11 Jun
This mind as wells be written in Japanese or cavemen, because I don't understand this.
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0 sats \ 1 reply \ @Cje95 11 Jun
I'm not sure what happened but I do not think you entered your bio right....
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0 sats \ 0 replies \ @k00b 11 Jun
Looks like a crawler/bot barfed or something.
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0 sats \ 0 replies \ @Johnson1s OP 11 Jun outlawed
stackers have outlawed this. turn on wild west mode in your /settings to see outlawed content.
0 sats \ 0 replies \ @Johnson1s OP 11 Jun outlawed
stackers have outlawed this. turn on wild west mode in your /settings to see outlawed content.