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	[mixcloud] Simplify url extraction
On the tracks I tested the server number in the url from the webpage is valid for the mp3 or the m4a file and any other number is invalid, it's a waste of time to check them.
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				@ -1,7 +1,6 @@
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from __future__ import unicode_literals
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import re
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import itertools
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from .common import InfoExtractor
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from ..compat import (
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@ -46,20 +45,16 @@ class MixcloudIE(InfoExtractor):
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        },
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    }]
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    def _get_url(self, track_id, template_url, server_number):
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        boundaries = (1, 30)
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        for nr in server_numbers(server_number, boundaries):
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            url = template_url % nr
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            try:
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                # We only want to know if the request succeed
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                # don't download the whole file
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                self._request_webpage(
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                    HEADRequest(url), track_id,
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                    'Checking URL %d/%d ...' % (nr, boundaries[-1]))
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                return url
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            except ExtractorError:
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                pass
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        return None
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    def _check_url(self, url, track_id, ext):
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        try:
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            # We only want to know if the request succeed
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            # don't download the whole file
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            self._request_webpage(
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                HEADRequest(url), track_id,
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                'Trying %s URL' % ext)
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            return True
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        except ExtractorError:
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            return False
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    def _real_extract(self, url):
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        mobj = re.match(self._VALID_URL, url)
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@ -72,15 +67,10 @@ class MixcloudIE(InfoExtractor):
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        preview_url = self._search_regex(
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            r'\s(?:data-preview-url|m-preview)="([^"]+)"', webpage, 'preview url')
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        song_url = preview_url.replace('/previews/', '/c/originals/')
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        server_number = int(self._search_regex(r'stream(\d+)', song_url, 'server number'))
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        template_url = re.sub(r'(stream\d*)', 'stream%d', song_url)
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        final_song_url = self._get_url(track_id, template_url, server_number)
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        if final_song_url is None:
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            self.to_screen('Trying with m4a extension')
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            template_url = template_url.replace('.mp3', '.m4a').replace('originals/', 'm4a/64/')
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            final_song_url = self._get_url(track_id, template_url, server_number)
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        if final_song_url is None:
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            raise ExtractorError('Unable to extract track url')
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        if not self._check_url(song_url, track_id, 'mp3'):
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            song_url = song_url.replace('.mp3', '.m4a').replace('originals/', 'm4a/64/')
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            if not self._check_url(song_url, track_id, 'm4a'):
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                raise ExtractorError('Unable to extract track url')
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        PREFIX = (
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            r'm-play-on-spacebar[^>]+'
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@ -107,7 +97,7 @@ class MixcloudIE(InfoExtractor):
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        return {
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            'id': track_id,
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            'title': title,
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            'url': final_song_url,
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            'url': song_url,
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            'description': description,
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            'thumbnail': thumbnail,
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            'uploader': uploader,
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@ -115,35 +105,3 @@ class MixcloudIE(InfoExtractor):
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            'view_count': view_count,
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            'like_count': like_count,
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        }
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def server_numbers(first, boundaries):
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    """ Server numbers to try in descending order of probable availability.
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    Starting from first (i.e. the number of the server hosting the preview file)
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    and going further and further up to the higher boundary and down to the
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    lower one in an alternating fashion. Namely:
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        server_numbers(2, (1, 5))
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        # Where the preview server is 2, min number is 1 and max is 5.
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        # Yields: 2, 3, 1, 4, 5
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    Why not random numbers or increasing sequences? Since from what I've seen,
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    full length files seem to be hosted on servers whose number is closer to
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    that of the preview; to be confirmed.
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    """
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    zip_longest = getattr(itertools, 'zip_longest', None)
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    if zip_longest is None:
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        # python 2.x
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        zip_longest = itertools.izip_longest
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    if len(boundaries) != 2:
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        raise ValueError("boundaries should be a two-element tuple")
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    min, max = boundaries
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    highs = range(first + 1, max + 1)
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    lows = range(first - 1, min - 1, -1)
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    rest = filter(
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        None, itertools.chain.from_iterable(zip_longest(highs, lows)))
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    yield first
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    for n in rest:
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        yield n
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