python - 如何通过YouTube JSON响应有效地分页？

我正在从YouTube channel 中收集有关视频的信息，该视频的数量大于50。

因此，这意味着我需要发出多个请求，因为每个JSON响应的最大结果是50个视频。

我找到了一些解决方案，现在代码看起来像这样

videoMetadata = [] #declaring our list, where the results will be stored

# First request
url = 'https://www.googleapis.com/youtube/v3/search?part=snippet&channelId='+CHANNEL_ID+'&maxResults=50&type=video&key='+API_KEY

response = urllib.request.urlopen(url) #makes the call to YouTube

videos = json.load(response) #decodes the response so we can work with it

nextPageToken = videos.get("nextPageToken") #gets the token of next page

# Retrieve all the rest of the pages

while nextPageToken:
    url = 'https://www.googleapis.com/youtube/v3/search?part=snippet&channelId='+CHANNEL_ID+'&maxResults=50&type=video&key='+API_KEY+"&pageToken="+nextPageToken
    response = urllib.request.urlopen(url)
    videos_next_page = json.load(response)
    nextPageToken = videos_next_page.get("nextPageToken")


# loops through results and appends it to videoMetadata list

# loop for the first page
for video in videos['items']:
    if video['id']['kind'] == 'youtube#video':
        videoMetadata.append(video['id']['videoId'])

# loop for the next page
for video in videos_next_page['items']:
    if video['id']['kind'] == 'youtube#video':

可以，但是也许有更好的解决方案，如何将多个JSON响应的结果存储在列表中？

将不胜感激任何建议。

最佳答案

实际上，除非您只有一个“下一页”，否则不会这样:

while nextPageToken:
    url = 'https://www.googleapis.com/youtube/v3/search?part=snippet&channelId='+CHANNEL_ID+'&maxResults=50&type=video&key='+API_KEY+"&pageToken="+nextPageToken
    response = urllib.request.urlopen(url)
    videos_next_page = json.load(response)
    nextPageToken = videos_next_page.get("nextPageToken")

每次迭代都会覆盖videos_next_page，因此您只会得到最后一页。



一旦取消序列化，“JSON响应的结果”就是普通的普通python对象(通常是dict)。然后将它们添加到列表中，就像处理其他任何事情一样。

这是一种可能的重写，可以正确处理此问题(并且也可以更好地利用内存)-警告:未经测试的代码，因此我不能保证没有任何错字或其他任何错误，但至少您可以理解。

def load_page(page_token=None):
    url = "https://www.googleapis.com/youtube/v3/search?part=snippet&channelId={}&maxResults=50&type=video&key={}".format(CHANNEL_ID, API_KEY)
    if page_token:
        url += ("&pageToken={}".format(page_token))
    response = urllib.request.urlopen(url) #makes the call to YouTube
    return json.load(response)

def collect_videos_meta(page):
    return [video['id']['videoId'] for video in page['items'] if video['id']['kind'] == 'youtube#video']

def main():
    videoMetadata = []
    nextPageToken = None # default initial value for the first page

    # using `while True` and `break` avoids having to repeat the same
    # code twice (once before the loop and once within the loop).
    # This is a very common pattern in Python, you just have to make
    # sure you will break out of the loop at some point...

    while True:
        page = load_page(nextPageToken)
        videoMetadata.extend(collect_videos_meta(page))
        nextPageToken = page.get("nextPageToken")
        if not nextPageToken:
            break

    # now do what you want with those data...
    print(videoMetadata)


if __name__ = "__main__":
    main()

关于python - 如何通过YouTube JSON响应有效地分页？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/57768954/