python:获取频道的所有 youtube 视频网址

2022-01-19 00:00:00 python youtube youtube-api

问题描述

我想获取特定频道的所有视频网址.我认为 json 与 python 或 java 将是一个不错的选择.我可以使用以下代码获取最新视频，但是如何获取所有视频链接 (>500)?

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

解决方案

将 max-results 从 1 增加到你想要的任意数量，但要注意他们不建议一次调用太多，并且会将你限制在 50 (https://developers.google.com/youtube/2.0/developers_guide_protocol_api_query_parameters).

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).

相反，您可以考虑以 25 个为一组抓取数据，例如，通过更改 start-index 直到没有返回.

Instead you could consider grabbing the data down in batches of 25, say, by changing the start-index until none came back.

这是我将如何做的代码

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

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