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Youtube Data API error: Subtitles are disabled for this video

Im trying to get the Title and transcript of all videos of a playlist:
from googleapiclient.discovery import build
from youtube_transcript_api import YouTubeTranscriptApi
import os
api_key = "*********************************"
#1.query API
rq = build("youtube", "v3", developerKey=api_key).playlistItems().list(
part="contentDetails, snippet",
playlistId="PL-osiE80TeTtoQCKZ03TU5fNfx2UY6U4p",
maxResults=50,
).execute()
#2.Create a list with video Ids and Titles
vid_ids = []
vid_title = []
for item in rq["items"]:
vid_ids.append(item["contentDetails"]["videoId"])
vid_title.append(item["snippet"]["title"])
#3.Get transcripts
srt = YouTubeTranscriptApi.get_transcripts(vid_ids)
print(srt)
But I get an error because one or more of those videos have no subtitles:
Could not retrieve a transcript for the video https://www.youtube.com/watch?v=D2lwk1Ukgz0! This is most likely caused by:
Subtitles are disabled for this video
What would you code in python to avoid this error, and get the transcripts of at least the rest of the videos of the playlist? maybe an If Statement (if the video has no subtitles jump to the next one) or similar?
Thanks in advance.

Try and except should help here:
for id in vid_ids:
try:
srt = YouTubeTranscriptApi.get_transcripts(id)
except:
print(f"{id} doesn't have a transcript")
This will basically ignore exceptions and tell you which id doesn't have a transcript.

Just try one video id at a time to get its transcript using try/except and pay attention not to pass directly a video id but instead an array of one video id to YouTubeTranscriptApi.get_transcripts otherwise it doesn't work.
So change:
#3.Get transcripts
srt = YouTubeTranscriptApi.get_transcripts(vid_ids)
print(srt)
For:
#3.Get transcripts
srt = []
for vid_id in vid_ids:
try:
srt += [YouTubeTranscriptApi.get_transcripts([vid_id])]
except:
srt += [({vid_id => []}, [])]
print(srt)

Mersive Solstive API: AttributeError: 'dict' object has no attribute 'm_displayInformation'

I have around 100 machines running Mersive Solstice, which is a wireless display tool. I'm trying to gather a few important pieces of information, in particular the fulfillment ID for the license for each installed instance.
Using the Solstice OpenControl API, found here, I whipped up a python script to grab everything I needed using a json GET. However, even when using the example GET from the documentation,
import requests
import json
url = ‘http://ip-of-machine/api/stats’
r = requests.get(url)
jsonStats = json.loads(r.text)
usersConnected = jsonStats.m_statistics.m_connectedUsers
I encounter:
Traceback (most recent call last):
File "C:/Python27/test.py", line 7, in <module>
usersConnected = jsonStats.m_statistics.m_connectedUsers
AttributeError: 'dict' object has no attribute 'm_statistics'
Which is very confusing. I've found plenty of similar questions on SO regarding this problem, but not one that's been specifically regarding wrong GET requests from the API Reference guide.
Additionally, here is my script:
import requests
import json
from time import sleep
url = 'test'
f = open("ip.txt", "r")
while(url != ""):
url = f.readline()
url = url.rstrip('\n')
print(url)
try:
r = requests.get(url)
except:
sleep(5)
jsonConfig = json.loads(r.text)
displayName = jsonConfig.m_displayInformation.m_displayName
hostName = jsonConfig.m_displayInformation.m_hostName
ipv4 = jsonConfig.m_displayInformation.m_ipv4
fulfillmentId = jsonConfig.m_licenseCuration.fulfillmentId
r.close()
f.close
I import the URL's from a text document for easy keeping. I'm able to make the connection to the /api/config JSON, and when the URL is put into a browser it does spit out the JSON records:

Json uses "Dicts" which are a type of array. You are just using them in the wrong way. I recommend reading Python Data Structures.
Json.Loads()
Returns a dictionary not a object. Do:
dict['key']['key']
Here is how your code should look:
import requests
import json
from time import sleep
url = 'test'
f = open("ip.txt", "r")
while(url != ""):
url = f.readline()
url = url.rstrip('\n')
print(url)
try:
response = requests.get(url)
json_object = json.loads(response .text)
displayName = json_object['m_displayInformation']['m_displayName']
hostName = json_object['m_displayInformation']['m_hostName']
ipv4 = json_object['m_displayInformation']['m_ipv4']
fulfillmentId = json_object['m_licenseCuration']['fulfillmentId']
except:
pass
response .close()
f.close()
I hope this was helpful!

Python 3 get song name from internet radio stream

How can I get song name from internet radio stream?
Python: Get name of shoutcast/internet radio station from url I looked here, but there is only getting name of radio station. But how to get name of the playing song? Here is stream link from where I want to get name of song. http://pool.cdn.lagardere.cz/fm-evropa2-128
How should I do it? Can you help me please?

To get the stream title, you need to request metadata. See shoutcast/icecast protocol description:
#!/usr/bin/env python
from __future__ import print_function
import re
import struct
import sys
try:
import urllib2
except ImportError: # Python 3
import urllib.request as urllib2
url = 'http://pool.cdn.lagardere.cz/fm-evropa2-128' # radio stream
encoding = 'latin1' # default: iso-8859-1 for mp3 and utf-8 for ogg streams
request = urllib2.Request(url, headers={'Icy-MetaData': 1}) # request metadata
response = urllib2.urlopen(request)
print(response.headers, file=sys.stderr)
metaint = int(response.headers['icy-metaint'])
for _ in range(10): # # title may be empty initially, try several times
response.read(metaint) # skip to metadata
metadata_length = struct.unpack('B', response.read(1))[0] * 16 # length byte
metadata = response.read(metadata_length).rstrip(b'\0')
print(metadata, file=sys.stderr)
# extract title from the metadata
m = re.search(br"StreamTitle='([^']*)';", metadata)
if m:
title = m.group(1)
if title:
break
else:
sys.exit('no title found')
print(title.decode(encoding, errors='replace'))
The stream title is empty in this case.

python: get all youtube video urls of a channel

I want to get all video url's of a specific channel. I think json with python or java would be a good choice. I can get the newest video with the following code, but how can I get ALL video links (>500)?
import urllib, json
author = 'Youtube_Username'
inp = urllib.urlopen(r'http://gdata.youtube.com/feeds/api/videos?max-results=1&alt=json&orderby=published&author=' + author)
resp = json.load(inp)
inp.close()
first = resp['feed']['entry'][0]
print first['title'] # video title
print first['link'][0]['href'] #url

After the youtube API change, max k.'s answer does not work. As a replacement, the function below provides a list of the youtube videos in a given channel. Please note that you need an API Key for it to work.
import urllib
import json
def get_all_video_in_channel(channel_id):
api_key = YOUR API KEY
base_video_url = 'https://www.youtube.com/watch?v='
base_search_url = 'https://www.googleapis.com/youtube/v3/search?'
first_url = base_search_url+'key={}&channelId={}&part=snippet,id&order=date&maxResults=25'.format(api_key, channel_id)
video_links = []
url = first_url
while True:
inp = urllib.urlopen(url)
resp = json.load(inp)
for i in resp['items']:
if i['id']['kind'] == "youtube#video":
video_links.append(base_video_url + i['id']['videoId'])
try:
next_page_token = resp['nextPageToken']
url = first_url + '&pageToken={}'.format(next_page_token)
except:
break
return video_links

Short answer:
Here's a library That can help with that.
pip install scrapetube
import scrapetube
videos = scrapetube.get_channel("UC9-y-6csu5WGm29I7JiwpnA")
for video in videos:
print(video['videoId'])
Long answer:
The module mentioned above was created by me due to a lack of any other solutions. Here's what i tried:
Selenium. It worked but had three big drawbacks: 1. It requires a web browser and driver to be installed. 2. has big CPU and memory requirements. 3. can't handle big channels.
Using youtube-dl. Like this:
import youtube_dl
youtube_dl_options = {
'skip_download': True,
'ignoreerrors': True
}
with youtube_dl.YoutubeDL(youtube_dl_options) as ydl:
videos = ydl.extract_info(f'https://www.youtube.com/channel/{channel_id}/videos')
This also works for small channels, but for bigger ones i would get blocked by youtube for making so many requests in such a short time (because youtube-dl downloads more info for every video in the channel).
So i made the library scrapetube which uses the web API to get all the videos.

Increase max-results from 1 to however many you want, but beware they don't advise grabbing too many in one call and will limit you at 50 (https://developers.google.com/youtube/2.0/developers_guide_protocol_api_query_parameters).
Instead you could consider grabbing the data down in batches of 25, say, by changing the start-index until none came back.
EDIT: Here's the code for how I would do it
import urllib, json
author = 'Youtube_Username'
foundAll = False
ind = 1
videos = []
while not foundAll:
inp = urllib.urlopen(r'http://gdata.youtube.com/feeds/api/videos?start-index={0}&max-results=50&alt=json&orderby=published&author={1}'.format( ind, author ) )
try:
resp = json.load(inp)
inp.close()
returnedVideos = resp['feed']['entry']
for video in returnedVideos:
videos.append( video )
ind += 50
print len( videos )
if ( len( returnedVideos ) < 50 ):
foundAll = True
except:
#catch the case where the number of videos in the channel is a multiple of 50
print "error"
foundAll = True
for video in videos:
print video['title'] # video title
print video['link'][0]['href'] #url

Based on the code found here and at some other places, I've written a small script that does this. My script uses v3 of Youtube's API and does not hit against the 500 results limit that Google has set for searches.
The code is available over at GitHub: https://github.com/dsebastien/youtubeChannelVideosFinder

Independent way of doing things. No api, no rate limit.
import requests
username = "marquesbrownlee"
url = "https://www.youtube.com/user/username/videos"
page = requests.get(url).content
data = str(page).split(' ')
item = 'href="/watch?'
vids = [line.replace('href="', 'youtube.com') for line in data if item in line] # list of all videos listed twice
print(vids[0]) # index the latest video
This above code will scrap only limited number of video url's max upto 60. How to grab all the videos url which is present in the channel. Can you please suggest.
This above code snippet will display only the list of all the videos which is listed twice. Not all the video url's in the channel.

Using Selenium Chrome Driver:
from selenium import webdriver
from webdriver_manager.chrome import ChromeDriverManager
import time
driverPath = ChromeDriverManager().install()
driver = webdriver.Chrome(driverPath)
url = 'https://www.youtube.com/howitshouldhaveended/videos'
driver.get(url)
height = driver.execute_script("return document.documentElement.scrollHeight")
previousHeight = -1
while previousHeight < height:
previousHeight = height
driver.execute_script(f'window.scrollTo(0,{height + 10000})')
time.sleep(1)
height = driver.execute_script("return document.documentElement.scrollHeight")
vidElements = driver.find_elements_by_id('thumbnail')
vid_urls = []
for v in vidElements:
vid_urls.append(v.get_attribute('href'))
This code has worked the few times I've tried it; however, you might need to tweak the sleep time, or add a way to recognize when the browser is still loading the extra information. It easily worked for me for getting a channel with 300+ videos, but it was having an issue with one that had 7000+ videos due to the time required to load the new videos on the browser becoming inconsistent.

I modified the script originally posted by dermasmid to fit my needs. This is the result:
import scrapetube
import sys
path = '_list.txt'
sys.stdout = open(path, 'w')
videos = scrapetube.get_channel("UC9-y-6csu5WGm29I7JiwpnA")
for video in videos:
print("https://www.youtube.com/watch?v="+str(video['videoId']))
# print(video['videoId'])
Basically it is saves all the URLs from the playlist into a "_list.txt" file. I am using this "_list.txt" file to download all the videos using the yt-dlp.exe. All the downloaded files have the .mp4 extension.
Now I do need to create another "_playlist.txt" file that contains all the FILENAMES coresponding to each URL from the "_List.txt".
For example, for: "https://www.youtube.com/watch?v=yG1m7oGZC48" to have "Apple M1 Ultra & NUMA - Computerphile.mp4" as output into the "_playlist.txt"

I do made some further improvements, to be able to add the channel URL into the console, print the result on screen and also into an external file called "_list.txt".
import scrapetube
import sys
path = '_list.txt'
print('**********************\n')
print("The result will be saved in '_list.txt' file.")
print("Enter Channel ID:")
# Prints the output in the console and into the '_list.txt' file.
class Logger:
def __init__(self, filename):
self.console = sys.stdout
self.file = open(filename, 'w')
def write(self, message):
self.console.write(message)
self.file.write(message)
def flush(self):
self.console.flush()
self.file.flush()
sys.stdout = Logger(path)
# Strip the: "https://www.youtube.com/channel/"
channel_id_input = input()
channel_id = channel_id_input.strip("https://www.youtube.com/channel/")
videos = scrapetube.get_channel(channel_id)
for video in videos:
print("https://www.youtube.com/watch?v="+str(video['videoId']))
# print(video['videoId'])

Invalid request URI while adding a video to playlist via youtube api

I have been unable to overcome this error while trying to add a video to my playlist using the youtube gdata python api.
gdata.service.RequestError: {'status':
400, 'body': 'Invalid request URI',
'reason': 'Bad Request'}
This seems to be the same error, but there are no solutions as yet. Any help guys?
import getpass
import gdata.youtube
import gdata.youtube.service
yt_service = gdata.youtube.service.YouTubeService()
# The YouTube API does not currently support HTTPS/SSL access.
yt_service.ssl = False
yt_service = gdata.youtube.service.YouTubeService()
yt_service.email = #myemail
yt_service.password = getpass.getpass()
yt_service.developer_key = #mykey
yt_service.source = #text
yt_service.client_id= #text
yt_service.ProgrammaticLogin()
feed = yt_service.GetYouTubePlaylistFeed(username='default')
# iterate through the feed as you would with any other
for entry in feed.entry:
if (entry.title.text == "test"):
lst = entry;
print entry.title.text, entry.id.text
custom_video_title = 'my test video on my test playlist'
custom_video_description = 'this is a test video on my test playlist'
video_id = 'Ncakifd_16k'
playlist_uri = lst.id.text
playlist_video_entry = yt_service.AddPlaylistVideoEntryToPlaylist(playlist_uri, video_id, custom_video_title, custom_video_description)
if isinstance(playlist_video_entry, gdata.youtube.YouTubePlaylistVideoEntry):
print 'Video added'
The confounding thing is that updating the playlist works, but adding a video does not.
playlist_entry_id = lst.id.text.split('/')[-1]
original_playlist_description = lst.description.text
updated_playlist = yt_service.UpdatePlaylist(playlist_entry_id,'test',original_playlist_description,playlist_private=False)
The video_id is not wrong because its the video from the sample code. What am I missing here? Somebody help!
Thanks.

Gdata seems to use v1 API. So, the relevant documentation is here: http://code.google.com/apis/youtube/1.0/developers_guide_protocol.html#Retrieving_a_playlist
This means, your "playlist_uri" should not take the value of "lst.id.text", but should take the "feedLink" element's "href" attribute in order to be used with "AddPlaylistVideoEntryToPlaylist"
Even if you happen to use v2 API, you should take the URI from the "content" element's "src" attribute as explained in the documentation, you get by substituting 2.0, in the above URL! (SO doesn't allow me to put two hyperlinks because i don't have enough reputations! :))