我们需要YouTube的频道名称的视频列表(使用API)。

我们可以通过下面的API获得一个频道列表(只有频道名):

https://gdata.youtube.com/feeds/api/channels?v=2&q=tendulkar

下面是频道的直接链接

https://www.youtube.com/channel/UCqAEtEr0A0Eo2IVcuWBfB9g

Or

WWW.YouTube.com/channel/HC-8jgBP-4rlI

现在,我们需要>> UCqAEtEr0A0Eo2IVcuWBfB9g或HC-8jgBP-4rlI频道的视频。

我们尝试

https://gdata.youtube.com/feeds/api/videos?v=2&uploader=partner&User=UC7Xayrf2k0NZiz3S04WuDNQ https://gdata.youtube.com/feeds/api/videos?v=2&uploader=partner&q=UC7Xayrf2k0NZiz3S04WuDNQ

但是,这并没有帮助。

我们需要把所有视频发到频道上。上传到一个频道的视频可以来自多个用户,因此我不认为提供用户参数会有帮助…


当前回答

使用gapi JavaScript API可以做到这一点

<script src="https://apis.google.com/js/api.js"></script>
const start = () => {
  gapi.client
    .init({
      apiKey: "your_youtubeApiKey",
      discoveryDocs: ["https://www.googleapis.com/discovery/v1/apis/youtube/v3/rest"],
      scope: "https://www.googleapis.com/auth/youtube.readonly",
    })
    .then(() => {
      console.log("gapi.client initiated");
    })
    .then(() =>
      gapi.client.youtube.channels.list({
        part: "snippet,contentDetails,statistics",
        id: "youtube_channelId",
        // forUsername: 'Bankless',
      })
    )
    .then(
      (res) =>
        // get the youtube related playlist id
        res.result.items[0].contentDetails.relatedPlaylists.uploads
    )
    .then((playlistId) =>
      gapi.client.youtube.playlistItems.list({
        part: "snippet",
        playlistId,
        maxResults: 50,
      })
    )
    .then((res) =>
      // get youtube videos snippets
      res.result.items.map((item) => item.snippet)
    )
    .then((snippets) =>
      snippets.map((snippet) => {
        const { title, description, resourceId } = snippet;
        const { videoId } = resourceId;
        return { title, description, videoId };
      })
    )
    .then((videos) => {
      console.log(videos);
    })
    .catch((err) => console.error(err));
};
gapi.load("client", start);

文档:

https://github.com/google/google-api-javascript-client https://developers.google.com/youtube/v3/guides/auth/client-side-web-apps#callinganapi

其他回答

使用gapi JavaScript API可以做到这一点

<script src="https://apis.google.com/js/api.js"></script>
const start = () => {
  gapi.client
    .init({
      apiKey: "your_youtubeApiKey",
      discoveryDocs: ["https://www.googleapis.com/discovery/v1/apis/youtube/v3/rest"],
      scope: "https://www.googleapis.com/auth/youtube.readonly",
    })
    .then(() => {
      console.log("gapi.client initiated");
    })
    .then(() =>
      gapi.client.youtube.channels.list({
        part: "snippet,contentDetails,statistics",
        id: "youtube_channelId",
        // forUsername: 'Bankless',
      })
    )
    .then(
      (res) =>
        // get the youtube related playlist id
        res.result.items[0].contentDetails.relatedPlaylists.uploads
    )
    .then((playlistId) =>
      gapi.client.youtube.playlistItems.list({
        part: "snippet",
        playlistId,
        maxResults: 50,
      })
    )
    .then((res) =>
      // get youtube videos snippets
      res.result.items.map((item) => item.snippet)
    )
    .then((snippets) =>
      snippets.map((snippet) => {
        const { title, description, resourceId } = snippet;
        const { videoId } = resourceId;
        return { title, description, videoId };
      })
    )
    .then((videos) => {
      console.log(videos);
    })
    .catch((err) => console.error(err));
};
gapi.load("client", start);

文档:

https://github.com/google/google-api-javascript-client https://developers.google.com/youtube/v3/guides/auth/client-side-web-apps#callinganapi

由于每个回答这个问题的人都有500个视频限制的问题,这里有一个使用Python 3中的youtube_dl的替代解决方案。此外,不需要API密钥。

安装youtube_dl: sudo pip3 Install youtube_dl 找出目标频道的频道id。ID将以UC开头。将Channel的C替换为Upload的U(即UU…),这是上传播放列表。 使用youtube-dl的播放列表下载功能。理想情况下,你不希望下载播放列表中的每个视频,这是默认的,而只是元数据。

示例(警告—需要数十分钟):

import youtube_dl, pickle

             # UCVTyTA7-g9nopHeHbeuvpRA is the channel id (1517+ videos)
PLAYLIST_ID = 'UUVTyTA7-g9nopHeHbeuvpRA'  # Late Night with Seth Meyers

with youtube_dl.YoutubeDL({'ignoreerrors': True}) as ydl:

    playd = ydl.extract_info(PLAYLIST_ID, download=False)

    with open('playlist.pickle', 'wb') as f:
        pickle.dump(playd, f, pickle.HIGHEST_PROTOCOL)

    vids = [vid for vid in playd['entries'] if 'A Closer Look' in vid['title']]
    print(sum('Trump' in vid['title'] for vid in vids), '/', len(vids))

获取频道列表:

通过forUserName获取频道列表:

https://www.googleapis.com/youtube/v3/channels?part=snippet,contentDetails,statistics&forUsername=Apple&key=

按频道id获取频道列表:

https://www.googleapis.com/youtube/v3/channels/?part=snippet,contentDetails,statistics&id=UCE_M8A5yxnLfW0KghEeajjw&key=

获取通道部分:

https://www.googleapis.com/youtube/v3/channelSections?part=snippet,contentDetails&channelId=UCE_M8A5yxnLfW0KghEeajjw&key=

获取播放列表:

按频道ID获取播放列表:

https://www.googleapis.com/youtube/v3/playlists?part=snippet,contentDetails&channelId=UCq-Fj5jknLsUf-MWSy4_brA&maxResults=50&key=

使用pageToken通过频道ID获取播放列表:

https://www.googleapis.com/youtube/v3/playlists?part=snippet,contentDetails&channelId=UCq-Fj5jknLsUf-MWSy4_brA&maxResults=50&key=&pageToken=CDIQAA

获取PlaylistItems:

通过PlayListId获取PlaylistItems列表:

https://www.googleapis.com/youtube/v3/playlistItems?part=snippet,contentDetails&maxResults=25&playlistId=PLHFlHpPjgk70Yv3kxQvkDEO5n5tMQia5I&key=

获取视频:

通过视频id获取视频列表:

https://www.googleapis.com/youtube/v3/videos?part=snippet,contentDetails,statistics&id=YxLCwfA1cLw&key=

通过多个视频id获取视频列表:

https://www.googleapis.com/youtube/v3/videos?part=snippet,contentDetails,statistics&id=YxLCwfA1cLw,Qgy6LaO3SB0,7yPJXGO2Dcw&key=

获取评论列表

按视频ID获取评论列表:

https://www.googleapis.com/youtube/v3/commentThreads?part=snippet replies&videoId = el * * * * kQak&key = ********** k

按频道ID获取评论列表:

https://www.googleapis.com/youtube/v3/commentThreads?part=snippet replies&channelId = U * * * * * q键= AI * * * * * * * * k

通过allThreadsRelatedToChannelId获取评论列表:

https://www.googleapis.com/youtube/v3/commentThreads?part=snippet replies&allThreadsRelatedToChannelId =加州大学* * * * * ntcQ&key = AI * * * * * k

这里所有的api都是Get方法。

基于频道id,我们不能直接得到所有的视频,这是这里的重点。

对于集成https://developers.google.com/youtube/v3/quickstart/ios?ver=swift

这是我的Python解决方案,使用谷歌API。 观察:

创建一个.env文件来存储API开发密钥,并将其放入.gitignore文件中 参数“forUserName”应该设置为Youtube频道的名称(用户名)。或者,您可以使用通道id,设置参数为“id”,而不是“forUserName”。 对象“playlistItem”让你访问每个视频。我只展示了它的标题,但还有很多其他属性。

import os
import googleapiclient.discovery
from decouple import config
def main():
    os.environ["OAUTHLIB_INSECURE_TRANSPORT"] = "1"

    api_service_name = "youtube"
    api_version = "v3"
    DEVELOPER_KEY = config('API_KEY')

    youtube = googleapiclient.discovery.build(
        api_service_name, api_version, developerKey = DEVELOPER_KEY)

    request = youtube.channels().list(
        part="contentDetails",
        forUsername="username",
        # id="oiwuereru8987",
    )

    response = request.execute()
    for item in response['items']:
        playlistId = item['contentDetails']['relatedPlaylists']['uploads']
        nextPageToken = ''
        while (nextPageToken != None):
            playlistResponse = youtube.playlistItems().list(
                part='snippet',
                playlistId=playlistId,
                maxResults=25,
                pageToken=nextPageToken
            )
            playlistResponse = playlistResponse.execute()
            print(playlistResponse.keys())
            for idx, playlistItem in enumerate(playlistResponse['items']):
                print(idx, playlistItem['snippet']['title'])
            if 'nextPageToken' in playlistResponse.keys():
                nextPageToken = playlistResponse['nextPageToken']
            else:
                nextPageToken = None

if __name__ == "__main__":
    main()

示例:.env文件

API_KEY=<Key_Here>

试着像下面这样做。它可能对你有帮助。

https://gdata.youtube.com/feeds/api/videos?author=cnn&v=2&orderby=updated&alt=jsonc&q=news

在这里作者,你可以指定你的频道名称和“q”,因为你可以给你的搜索关键字。