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Clean up markdown parsing since tuples aren't used anymore

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Lonami Exo 2017-11-10 11:41:49 +01:00
parent 7d75eebdab
commit cb3f20db65
1 changed files with 36 additions and 32 deletions
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68
telethon/extensions/markdown.py
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@ -48,12 +48,12 @@ def parse(message, delimiters=None, url_re=None):
i = 0
result = []
current = None
end_delimiter = None
# Work on byte level with the utf-16le encoding to get the offsets right.
# The offset will just be half the index we're at.
message = message.encode('utf-16le')
while i < len(message):
url_match = None
if url_re and current is None:
# If we're not inside a previous match since Telegram doesn't allow
# nested message entities, try matching the URL from the i'th pos.
@ -70,42 +70,46 @@ def parse(message, delimiters=None, url_re=None):
offset=i // 2, length=len(url_match.group(1)) // 2,
url=url_match.group(2).decode('utf-16le')
))
# We matched the delimiter which is now gone, and we'll add
# +2 before next iteration which will make us skip a character.
# Go back by one utf-16 encoded character (-2) to avoid it.
i += len(url_match.group(1)) - 2
i += len(url_match.group(1))
# Next loop iteration, don't check delimiters, since
# a new inline URL might be right after this one.
continue
if not url_match:
if end_delimiter is None:
# We're not expecting any delimiter, so check them all
for d, m in delimiters.items():
# Slice the string at the current i'th position to see if
# it matches the current delimiter d.
if message[i:i + len(d)] == d:
if current is not None and not isinstance(current, m):
# We were inside another delimiter/mode, ignore this.
continue
# it matches the current delimiter d, otherwise skip it.
if message[i:i + len(d)] != d:
continue
if message[i + len(d):i + 2 * len(d)] == d:
# The same delimiter can't be right afterwards, if
# this were the case we would match empty strings
# like `` which we don't want to.
continue
if message[i + len(d):i + 2 * len(d)] == d:
# The same delimiter can't be right afterwards, if
# this were the case we would match empty strings
# like `` which we don't want to.
continue
# Get rid of the delimiter by slicing it away
message = message[:i] + message[i + len(d):]
if current is None:
if m == MessageEntityPre:
# Special case, also has 'lang'
current = MessageEntityPre(i // 2, None, '')
else:
current = m(i // 2, None)
# No need to i -= 2 here because it's been already
# checked that next character won't be a delimiter.
else:
current.length = (i // 2) - current.offset
result.append(current)
current = None
i -= 2 # Delimiter matched and gone, go back 1 char
break
# Get rid of the delimiter by slicing it away
message = message[:i] + message[i + len(d):]
if m == MessageEntityPre:
# Special case, also has 'lang'
current = m(i // 2, None, '')
else:
current = m(i // 2, None)
end_delimiter = d # We expect the same delimiter.
break
elif message[i:i + len(end_delimiter)] == end_delimiter:
message = message[:i] + message[i + len(end_delimiter):]
current.length = (i // 2) - current.offset
result.append(current)
current, end_delimiter = None, None
# Don't increment i here as we matched a delimiter,
# and there may be a new one right after. This is
# different than when encountering the first delimiter,
# as we already know there won't be the same right after.
continue
# Next iteration, utf-16 encoded characters need 2 bytes.
i += 2