Python tornado.util.re_unescape() Examples

The following are 7 code examples of tornado.util.re_unescape(). You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. You may also want to check out all available functions/classes of the module tornado.util , or try the search function .
Example #1
Source File: routing.py    From opendevops with GNU General Public License v3.0 5 votes vote down vote up 
def _find_groups(self) -> Tuple[Optional[str], Optional[int]]:
        """Returns a tuple (reverse string, group count) for a url.

        For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
        would return ('/%s/%s/', 2).
        """
        pattern = self.regex.pattern
        if pattern.startswith("^"):
            pattern = pattern[1:]
        if pattern.endswith("$"):
            pattern = pattern[:-1]

        if self.regex.groups != pattern.count("("):
            # The pattern is too complicated for our simplistic matching,
            # so we can't support reversing it.
            return None, None

        pieces = []
        for fragment in pattern.split("("):
            if ")" in fragment:
                paren_loc = fragment.index(")")
                if paren_loc >= 0:
                    pieces.append("%s" + fragment[paren_loc + 1 :])
            else:
                try:
                    unescaped_fragment = re_unescape(fragment)
                except ValueError:
                    # If we can't unescape part of it, we can't
                    # reverse this url.
                    return (None, None)
                pieces.append(unescaped_fragment)

        return "".join(pieces), self.regex.groups
Example #2
Source File: views.py    From tornado-swirl with MIT License 5 votes vote down vote up 
def _find_groups(url):
    """Returns a tuple (reverse string, group count) for a url.

    For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
    would return ('/%s/%s/', 2).
    """
    regex = re.compile(url)
    pattern = url
    if pattern.startswith('^'):
        pattern = pattern[1:]
    if pattern.endswith('$'):
        pattern = pattern[:-1]

    if regex.groups != pattern.count('('):
        # The pattern is too complicated for our simplistic matching,
        # so we can't support reversing it.
        return None, None

    pieces = []
    for fragment in pattern.split('('):
        if ')' in fragment:
            paren_loc = fragment.index(')')
            if paren_loc >= 0:
                pieces.append('%s' + fragment[paren_loc + 1:])
        else:
            try:
                unescaped_fragment = re_unescape(fragment)
            except ValueError:
                # If we can't unescape part of it, we can't
                # reverse this url.
                return (None, None)
            pieces.append(unescaped_fragment)
    return ''.join(pieces), regex.groups
Example #3
Source File: routing.py    From teleport with Apache License 2.0 5 votes vote down vote up 
def _find_groups(self):
        """Returns a tuple (reverse string, group count) for a url.

        For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
        would return ('/%s/%s/', 2).
        """
        pattern = self.regex.pattern
        if pattern.startswith('^'):
            pattern = pattern[1:]
        if pattern.endswith('$'):
            pattern = pattern[:-1]

        if self.regex.groups != pattern.count('('):
            # The pattern is too complicated for our simplistic matching,
            # so we can't support reversing it.
            return None, None

        pieces = []
        for fragment in pattern.split('('):
            if ')' in fragment:
                paren_loc = fragment.index(')')
                if paren_loc >= 0:
                    pieces.append('%s' + fragment[paren_loc + 1:])
            else:
                try:
                    unescaped_fragment = re_unescape(fragment)
                except ValueError:
                    # If we can't unescape part of it, we can't
                    # reverse this url.
                    return (None, None)
                pieces.append(unescaped_fragment)

        return ''.join(pieces), self.regex.groups
Example #4
Source File: routing.py    From teleport with Apache License 2.0 5 votes vote down vote up 
def _find_groups(self) -> Tuple[Optional[str], Optional[int]]:
        """Returns a tuple (reverse string, group count) for a url.

        For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
        would return ('/%s/%s/', 2).
        """
        pattern = self.regex.pattern
        if pattern.startswith("^"):
            pattern = pattern[1:]
        if pattern.endswith("$"):
            pattern = pattern[:-1]

        if self.regex.groups != pattern.count("("):
            # The pattern is too complicated for our simplistic matching,
            # so we can't support reversing it.
            return None, None

        pieces = []
        for fragment in pattern.split("("):
            if ")" in fragment:
                paren_loc = fragment.index(")")
                if paren_loc >= 0:
                    pieces.append("%s" + fragment[paren_loc + 1 :])
            else:
                try:
                    unescaped_fragment = re_unescape(fragment)
                except ValueError:
                    # If we can't unescape part of it, we can't
                    # reverse this url.
                    return (None, None)
                pieces.append(unescaped_fragment)

        return "".join(pieces), self.regex.groups
Example #5
Source File: routing.py    From teleport with Apache License 2.0 5 votes vote down vote up 
def _find_groups(self) -> Tuple[Optional[str], Optional[int]]:
        """Returns a tuple (reverse string, group count) for a url.

        For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
        would return ('/%s/%s/', 2).
        """
        pattern = self.regex.pattern
        if pattern.startswith("^"):
            pattern = pattern[1:]
        if pattern.endswith("$"):
            pattern = pattern[:-1]

        if self.regex.groups != pattern.count("("):
            # The pattern is too complicated for our simplistic matching,
            # so we can't support reversing it.
            return None, None

        pieces = []
        for fragment in pattern.split("("):
            if ")" in fragment:
                paren_loc = fragment.index(")")
                if paren_loc >= 0:
                    pieces.append("%s" + fragment[paren_loc + 1 :])
            else:
                try:
                    unescaped_fragment = re_unescape(fragment)
                except ValueError:
                    # If we can't unescape part of it, we can't
                    # reverse this url.
                    return (None, None)
                pieces.append(unescaped_fragment)

        return "".join(pieces), self.regex.groups
Example #6
Source File: routing.py    From pySINDy with MIT License 5 votes vote down vote up 
def _find_groups(self):
        """Returns a tuple (reverse string, group count) for a url.

        For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
        would return ('/%s/%s/', 2).
        """
        pattern = self.regex.pattern
        if pattern.startswith('^'):
            pattern = pattern[1:]
        if pattern.endswith('$'):
            pattern = pattern[:-1]

        if self.regex.groups != pattern.count('('):
            # The pattern is too complicated for our simplistic matching,
            # so we can't support reversing it.
            return None, None

        pieces = []
        for fragment in pattern.split('('):
            if ')' in fragment:
                paren_loc = fragment.index(')')
                if paren_loc >= 0:
                    pieces.append('%s' + fragment[paren_loc + 1:])
            else:
                try:
                    unescaped_fragment = re_unescape(fragment)
                except ValueError:
                    # If we can't unescape part of it, we can't
                    # reverse this url.
                    return (None, None)
                pieces.append(unescaped_fragment)

        return ''.join(pieces), self.regex.groups
Example #7
Source File: routing.py    From V1EngineeringInc-Docs with Creative Commons Attribution Share Alike 4.0 International 5 votes vote down vote up 
def _find_groups(self) -> Tuple[Optional[str], Optional[int]]:
        """Returns a tuple (reverse string, group count) for a url.

        For example: Given the url pattern /([0-9]{4})/([a-z-]+)/, this method
        would return ('/%s/%s/', 2).
        """
        pattern = self.regex.pattern
        if pattern.startswith("^"):
            pattern = pattern[1:]
        if pattern.endswith("$"):
            pattern = pattern[:-1]

        if self.regex.groups != pattern.count("("):
            # The pattern is too complicated for our simplistic matching,
            # so we can't support reversing it.
            return None, None

        pieces = []
        for fragment in pattern.split("("):
            if ")" in fragment:
                paren_loc = fragment.index(")")
                if paren_loc >= 0:
                    pieces.append("%s" + fragment[paren_loc + 1 :])
            else:
                try:
                    unescaped_fragment = re_unescape(fragment)
                except ValueError:
                    # If we can't unescape part of it, we can't
                    # reverse this url.
                    return (None, None)
                pieces.append(unescaped_fragment)

        return "".join(pieces), self.regex.groups