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# coding: utf-8

from __future__ import unicode_literals

from functools import partial

import re

from bpython.lazyre import LazyReCompile

from curtsies.termformatconstants import FG_COLORS, BG_COLORS, colors

from curtsies.formatstring import fmtstr, FmtStr

cnames = dict(zip('krgybmcwd', colors + ('default',)))

def func_for_letter(l, default='k'):

"""Returns FmtStr constructor for a bpython-style color code"""

if l == 'd':

l = default

elif l == 'D':

l = default.upper()

return partial(fmtstr, fg=cnames[l.lower()], bold=l.isupper())

def color_for_letter(l, default='k'):

if l == 'd':

l = default

return cnames[l.lower()]

def parse(s):

"""Returns a FmtStr object from a bpython-formatted colored string"""

rest = s

stuff = []

while True:

if not rest:

break

start, rest = peel_off_string(rest)

stuff.append(start)

return (sum((fs_from_match(d) for d in stuff[1:]), fs_from_match(stuff[0]))

if len(stuff) > 0

else FmtStr())

def fs_from_match(d):

atts = {}

if d['fg']:

# this isn't according to spec as I understand it

if d['fg'].isupper():

d['bold'] = True

# TODO figure out why boldness isn't based on presence of \x02

color = cnames[d['fg'].lower()]

if color != 'default':

atts['fg'] = FG_COLORS[color]

if d['bg']:

if d['bg'] == 'I':

# hack for finding the "inverse"

color = colors[(colors.index(color) + (len(colors) // 2)) %

len(colors)]

else:

color = cnames[d['bg'].lower()]

if color != 'default':

atts['bg'] = BG_COLORS[color]

if d['bold']:

atts['bold'] = True

return fmtstr(d['string'], **atts)

peel_off_string_re = LazyReCompile(

r"""(?P<colormarker>\x01

(?P<fg>[krgybmcwdKRGYBMCWD]?)

(?P<bg>[krgybmcwdKRGYBMCWDI]?)?)

(?P<bold>\x02?)

\x03

(?P<string>[^\x04]*)

\x04

(?P<rest>.*)

""", re.VERBOSE | re.DOTALL)

def peel_off_string(s):

m = peel_off_string_re.match(s)

assert m, repr(s)

d = m.groupdict()

rest = d['rest']

del d['rest']

return d, rest