I am having trouble getting a video entry which includes a link rel="edit". I need such an entry in order to be able to call DeleteVideoEntry(...) on it.
I am retrieving the video using GetYouTubeVideoEntry(youtube_id=XXXXXXX). My yt_service is initialized with a username, password, and a developer key. I use ProgrammaticLogin. This part seems to work fine. I use the same yt_service to upload said video earlier. Also, if I change the developer key to something bogus (during debugging) and try to authenticate, I get a 403 error. This leads me to believe that authentication works OK.
Needsless to say, the video entry retrieved with GetYouTubeVideoEntry(youtube_id=XXXXXXX) does not contain the edit link and I cannot use the entry in a DeleteVideoEntry(...) call.
Is there some special way to get a video entry which will contain a link element with a rel="edit"? Can anyone suggest some way to resolve my issue? Could this possibly be a bug?
Update:
For the records, when I tried getting the feed of all my uploads, and then looping through the video entries, the video entries do have an edit link. So using this works:
uri = 'http://gdata.youtube.com/feeds/api/users/%s/uploads' % username
feed = yt_service.GetYouTubeVideoFeed(uri)
for entry in feed.entry:
yt_service.DeleteVideoEntry(entry)
But this does not:
entry = yt_service.GetYouTubeVideoEntry(video_id = video.youtube_id)
yt_service.DeleteVideoEntry(entry)
Using the same yt_service.
I've just deleted youtube video using gdata and ProgrammaticLogin()
Here is some steps to reproduce:
import gdata.youtube.service
yt_service = gdata.youtube.service.YouTubeService()
yt_service.developer_key = 'developer_key'
yt_service.email = 'email'
yt_service.password = 'password'
yt_service.ProgrammaticLogin()
# video_id should looks like 'iu6Gq-tUsTc'
uri = 'https://gdata.youtube.com/feeds/api/users/%s/uploads/%s' % (username, video_id)
entry = yt_service.GetYouTubeUserEntry(uri=uri)
response = yt_service.DeleteVideoEntry(entry)
print response # True
yt_service.GetYouTubeVideoFeed(uri) works because GetYouTubeVideoFeed doesn't check uri and just calls self.Get(uri, ...) but originaly, I think, it expected 'https://gdata.youtube.com/feeds/api/videos' uri.
vice versa yt_service.GetYouTubeVideoEntry() use YOUTUBE_VIDEO_URI = 'https://gdata.youtube.com/feeds/api/videos' but this entry doesn't contains rel="edit"
Hope that helps you out
You can view the HTTP headers of the generated requests by setting the debug flag to true. This is as simple as:
yt_service = gdata.youtube.service.YouTubeService()
yt_service.debug = True
You can read about this in the documentation here.
Related
I am trying to use requests (python) to grab some pages from a website that requires me to be logged in.
I did inspect the login page to check out the username and password headers. But I found the names for those fields are not the standard 'username', 'password' used by most sites as you can see from the below screenshots
password field
I used them that way in my python script but each time I get a 'wrong syntax' error. Even sublimetext displayed a part of the name in orange as you can see from the pix below
From this I know there must be some problem with the name. But try to escape the $ signs did not help.
Even the login.aspx header disappears before google chrome could register it on the network.
The site is www dot bncnetwork dot net
I'd be happy if someone could help me figure out what to do about this.
Here is the code`import requests
import requests
def get_project_page(seed_page):
username = "*******************"
password = "*******************"
bnc_login = dict(ctl00$MainContent$txtEmailID=username, ctl00$MainContent$txtPassword=password)
sess_req = requests.Session()
sess_req.get(seed_page)
sess_req.post(seed_page, data=bnc_login, headers={"Referer":"http://www.bncnetwork.net/MyBNC.aspx"})
page = sess_req.get(seed_page)
return page.text`
You need to use strings for the keys, the $ will cause a syntax error if you don't:
data = {"ctl00$MainContent$txtPassword":password, "ctl00$MainContent$txtEmailID":email}
There are evenvalidation fileds etc.. to be filled in also, follow the logic from this answer to fill them out, all the fields can be seen in chrome tools:
My goal is to connect to Youtube API and download the URLs of specific music producers.I found the following script which I used from the following link: https://www.youtube.com/watch?v=_M_wle0Iq9M. In the video the code works beautifully. But when I try it on python 2.7 it gives me KeyError:'items'.
I know KeyErrors can occur when there is an incorrect use of a dictionary or when a key doesn't exist.
I have tried going to the google developers site for youtube to make sure that 'items' exist and it does.
I am also aware that using get() may be helpful for my problem but I am not sure. Any suggestions to fixing my KeyError using the following code or any suggestions on how to improve my code to reach my main goal of downloading the URLs (I have a Youtube API)?
Here is the code:
#these modules help with HTTP request from Youtube
import urllib
import urllib2
import json
API_KEY = open("/Users/ereyes/Desktop/APIKey.rtf","r")
API_KEY = API_KEY.read()
searchTerm = raw_input('Search for a video:')
searchTerm = urllib.quote_plus(searchTerm)
url = 'https://www.googleapis.com/youtube/v3/search?part=snippet&q='+searchTerm+'&key='+API_KEY
response = urllib.urlopen(url)
videos = json.load(response)
videoMetadata = [] #declaring our list
for video in videos['items']: #"for loop" cycle through json response and searches in items
if video['id']['kind'] == 'youtube#video': #makes sure that item we are looking at is only videos
videoMetadata.append(video['snippet']['title']+ # getting title of video and putting into list
"\nhttp://youtube.com/watch?v="+video['id']['videoId'])
videoMetadata.sort(); # sorts our list alphaetically
print ("\nSearch Results:\n") #print out search results
for metadata in videoMetadata:
print (metadata)+"\n"
raw_input('Press Enter to Exit')
The problem is most likely a combination of using an RTF file instead of a plain text file for the API key and you seem to be confused whether to use urllib or urllib2 since you imported both.
Personally, I would recommend requests, but I think you need to read() the contents of the request to get a string
response = urllib.urlopen(url).read()
You can check that by printing the response variable
I am trying to use PRAW to get new posts from subreddits on Reddit. The following code snippet shows how I get new items on a particular subreddit.
Is there a way to also get the URL of the particular submission?
submissions = r.get_subreddit('todayilearned')
submission = submissions.get_new(limit=1)
sub = [str(x) for x in submission]
print sub
PRAW allows you to do this:
To get the submitted link you can use submission.url
[submission] = submissions.get_new(limit=1)
print submission.url
Or if you're looking for the URL for the actual post to Reddit then you can use permalink
[submission] = submissions.get_new(limit=1)
print submission.permalink
The documentation lists a short_link property that returns a shortened version of the url to the submission. It does not appear that the full url is similarly provided, though it seems that it could be reconstructed from the subreddit name and the submission's id, which is stored in submission.id.
In summary, use:
[submission] = submissions.get_new(limit=1)
submission.short_link
to get a link to the submission.
I'm trying to crawl Youtube to retrieve information about a group of users (approx. 200 people). I'm interested in looking for relationships between the users:
contacts
subscribers
subscriptions
what videos they commented on
etc
I've managed to get contact information with the following source:
import gdata.youtube
import gdata.youtube.service
from gdata.service import RequestError
from pub_author import KEY, NAME_REGEX
def get_details(name):
yt_service = gdata.youtube.service.YouTubeService()
yt_service.developer_key = KEY
contact_feed = yt_service.GetYouTubeContactFeed(username=name)
contacts = [ e.title.text for e in contact_feed.entry ]
return contacts
I can't seem the get the other bits of information I need. The reference guide says that I can grab the XML feed from http://gdata.youtube.com/feeds/api/users/username/subscriptions?v=2 (for some arbitrary user). However, if I try to get other users' subscriptions, I get the a 403 error with the following message:
User must be logged in to access these subscriptions.
If I use the gdata API:
sub_feed = yt_service.GetYouTubeSubscriptionFeed(username=name)
sub = [ e.title.text for e in contact_feed.entry ]
then I get the same error.
How can I get these subscriptions without logging in? It should be possible, as you can access this information without logging in to the Youtube web-site.
Also, there seems to be no feed for the subscribers of particular user. Is this information available through the API?
EDIT
So, it appears this can't be done through the API. I had to do this the quick and dirty way:
for f in `cat users.txt`; do wget "www.youtube.com/profile?user=$f&view=subscriptions" --output-document subscriptions/$f.html; done
Then use this script to get out the usernames from the downloaded HTML files:
"""Extract usernames from a Youtube profile using regex"""
import re
def main():
import sys
lines = open(sys.argv[1]).read().split('\n')
#
# The html files has two <a href="..."> tags for each user: once for an
# image thumbnail, and once for a text link.
#
users = set()
for l in lines:
match = re.search('<a href="/user/(?P<name>[^"]+)" onmousedown', l)
if match:
users.add(match.group('name'))
users = list(users)
users.sort()
print users
if __name__ == '__main__':
main()
In order to access a user's subscriptions feed without the user being logged in, the user must check the "Subscribe to a channel" checkbox under his Account Sharing settings.
Currently, there is no direct way to get a channel's subscribers through the gdata API. In fact, there has been an outstanding feature request for it that has remained open for over 3 years! See Retrieving a list of a user's subscribers?.
I'm currently trying to get a grasp on pycurl. I'm attempting to login to a website. After logging into the site it should redirect to the main page. However when trying this script it just gets returned to the login page. What might I be doing wrong?
import pycurl
import urllib
import StringIO
pf = {'username' : 'user', 'password' : 'pass' }
fields = urllib.urlencode(pf)
pageContents = StringIO.StringIO()
p = pycurl.Curl()
p.setopt(pycurl.FOLLOWLOCATION, 1)
p.setopt(pycurl.COOKIEFILE, './cookie_test.txt')
p.setopt(pycurl.COOKIEJAR, './cookie_test.txt')
p.setopt(pycurl.POST, 1)
p.setopt(pycurl.POSTFIELDS, fields)
p.setopt(pycurl.WRITEFUNCTION, pageContents.write)
p.setopt(pycurl.URL, 'http://localhost')
p.perform()
pageContents.seek(0)
print pageContents.readlines()
EDIT: As pointed out by Peter the URL should point to a login URL but the site I'm trying to get this to work for fails to show me what URL this would be. The form's action just points to the home page ( /index.html )
As you're troubleshooting this problem, I suggest getting a browser plugin like FireBug or LiveHTTPHeaders (I suggest Firefox plugins, but there are similar plugins for other browsers as well). Then you can exercise a request to the site and see what action (URL), method, and form parameters are being passed to the target server. This will likely help elucidate the crux of the problem.
If that's no help, you may consider using a different tool for your mechanization. I've used ClientForm and BeautifulSoup to perform similar operations. Based on what I've read in the pycURL docs and your code above, ClientForm might be a better tool to use. ClientForm will parse your HTML page, locate the forms on it (including login forms), and construct the appropriate request for you based on the answers you supply to the form. You could even use ClientForm with pycURL... but at least ClientForm will provide you with the appropriate action to which to POST, and construct all of the appropriate parameters.
Be aware, though, that if there is JavaScript handling any necessary part of the login form, even ClientForm can't help you there. You will need something that interprets the JavaScript to effectively automate the login. In that case, I've used SeleniumRC to control a browser (and I let the browser handle the JavaScript).
One of the golden rule, you need to 'brake the ice', have debugging enabled when trying to solve pycurl example:
Note: don't forget to use p.close() after p.perform()
def test(debug_type, debug_msg):
if len(debug_msg) < 300:
print "debug(%d): %s" % (debug_type, debug_msg.strip())
p.setopt(pycurl.VERBOSE, True)
p.setopt(pycurl.DEBUGFUNCTION, test)
Now you can see how your code is breathing, because you have debugging enabled
import pycurl
import urllib
import StringIO
def test(debug_type, debug_msg):
if len(debug_msg) < 300:
print "debug(%d): %s" % (debug_type, debug_msg.strip())
pf = {'username' : 'user', 'password' : 'pass' }
fields = urllib.urlencode(pf)
pageContents = StringIO.StringIO()
p = pycurl.Curl()
p.setopt(pycurl.FOLLOWLOCATION, 1)
p.setopt(pycurl.COOKIEFILE, './cookie_test.txt')
p.setopt(pycurl.COOKIEJAR, './cookie_test.txt')
p.setopt(pycurl.POST, 1)
p.setopt(pycurl.POSTFIELDS, fields)
p.setopt(pycurl.WRITEFUNCTION, pageContents.write)
p.setopt(pycurl.VERBOSE, True)
p.setopt(pycurl.DEBUGFUNCTION, test)
p.setopt(pycurl.URL, 'http://localhost')
p.perform()
p.close() # This is mandatory.
pageContents.seek(0)
print pageContents.readlines()