How can I extract video ID from YouTube's link in Python?

20,212

Solution 1

Python has a library for parsing URLs.

import urlparse
url_data = urlparse.urlparse("http://www.youtube.com/watch?v=z_AbfPXTKms&NR=1")
query = urlparse.parse_qs(url_data.query)
video = query["v"][0]

Solution 2

I've created youtube id parser without regexp:

import urlparse

def video_id(value):
    """
    Examples:
    - http://youtu.be/SA2iWivDJiE
    - http://www.youtube.com/watch?v=_oPAwA_Udwc&feature=feedu
    - http://www.youtube.com/embed/SA2iWivDJiE
    - http://www.youtube.com/v/SA2iWivDJiE?version=3&hl=en_US
    """
    query = urlparse.urlparse(value)
    if query.hostname == 'youtu.be':
        return query.path[1:]
    if query.hostname in ('www.youtube.com', 'youtube.com'):
        if query.path == '/watch':
            p = urlparse.parse_qs(query.query)
            return p['v'][0]
        if query.path[:7] == '/embed/':
            return query.path.split('/')[2]
        if query.path[:3] == '/v/':
            return query.path.split('/')[2]
    # fail?
    return None

Solution 3

This is the Python3 version of Mikhail Kashkin's solution with added scenarios.

from urllib.parse import urlparse, parse_qs
from contextlib import suppress


# noinspection PyTypeChecker
def get_yt_id(url, ignore_playlist=False):
    # Examples:
    # - http://youtu.be/SA2iWivDJiE
    # - http://www.youtube.com/watch?v=_oPAwA_Udwc&feature=feedu
    # - http://www.youtube.com/embed/SA2iWivDJiE
    # - http://www.youtube.com/v/SA2iWivDJiE?version=3&hl=en_US
    query = urlparse(url)
    if query.hostname == 'youtu.be': return query.path[1:]
    if query.hostname in {'www.youtube.com', 'youtube.com', 'music.youtube.com'}:
        if not ignore_playlist:
        # use case: get playlist id not current video in playlist
            with suppress(KeyError):
                return parse_qs(query.query)['list'][0]
        if query.path == '/watch': return parse_qs(query.query)['v'][0]
        if query.path[:7] == '/watch/': return query.path.split('/')[1]
        if query.path[:7] == '/embed/': return query.path.split('/')[2]
        if query.path[:3] == '/v/': return query.path.split('/')[2]
   # returns None for invalid YouTube url

Solution 4

Here is RegExp it cover these cases enter image description here

((?<=(v|V)/)|(?<=be/)|(?<=(\?|\&)v=)|(?<=embed/))([\w-]+)

Solution 5

match = re.search(r"youtube\.com/.*v=([^&]*)", "http://www.youtube.com/watch?v=z_AbfPXTKms&test=123")
if match:
    result = match.group(1)
else:
    result = ""

Untested.

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20,212
decarbo
Author by

decarbo

Updated on January 05, 2022

Comments

  • decarbo
    decarbo over 2 years

    I know this can be easily done using PHP's parse_url and parse_str functions:

    $subject = "http://www.youtube.com/watch?v=z_AbfPXTKms&NR=1";
    $url = parse_url($subject);
    parse_str($url['query'], $query);
    var_dump($query);
    

    But how to achieve this using Python? I can do urlparse but what next?