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[tools] Fixed failure in syntax highlighting.

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- If the theme doesn't know a token type use a default style as a
  best effort.
This commit is contained in:
Andrew Hamilton 2017-07-11 10:59:27 +01:00
parent ed5164c761
commit 75c7db0f17
1 changed files with 8 additions and 3 deletions
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11
vigil/tools.py
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@ -148,8 +148,12 @@ def _syntax_highlight(text, lexer, style):
hex_rgb = hex_rgb[1:] hex_rgb = hex_rgb[1:]
return tuple(eval("0x"+hex_rgb[index:index+2]) for index in [0, 2, 4]) return tuple(eval("0x"+hex_rgb[index:index+2]) for index in [0, 2, 4])
def _char_style_for_token_type(token_type, default_bg_color): def _char_style_for_token_type(token_type, default_bg_color,
token_style = style.style_for_token(token_type) default_style):
try:
token_style = style.style_for_token(token_type)
except KeyError:
return default_style
fg_color = (termstr.Color.black if token_style["color"] is None fg_color = (termstr.Color.black if token_style["color"] is None
else _parse_rgb(token_style["color"])) else _parse_rgb(token_style["color"]))
bg_color = (default_bg_color if token_style["bgcolor"] is None bg_color = (default_bg_color if token_style["bgcolor"] is None
@ -158,9 +162,10 @@ def _syntax_highlight(text, lexer, style):
token_style["italic"], token_style["italic"],
token_style["underline"]) token_style["underline"])
default_bg_color = _parse_rgb(style.background_color) default_bg_color = _parse_rgb(style.background_color)
default_style = termstr.CharStyle(bg_color=default_bg_color)
text = fill3.join( text = fill3.join(
"", [termstr.TermStr(text, _char_style_for_token_type( "", [termstr.TermStr(text, _char_style_for_token_type(
token_type, default_bg_color)) token_type, default_bg_color, default_style))
for token_type, text in pygments.lex(text, lexer)]) for token_type, text in pygments.lex(text, lexer)])
return fill3.Text(text, pad_char=termstr.TermStr(" ").bg_color( return fill3.Text(text, pad_char=termstr.TermStr(" ").bg_color(
default_bg_color)) default_bg_color))