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…
最近,我不得不从一个频道检索所有视频,根据YouTube开发者文档:
https://developers.google.com/youtube/v3/docs/playlistItems/list
function playlistItemsListByPlaylistId($service, $part, $params) {
$params = array_filter($params);
$response = $service->playlistItems->listPlaylistItems(
$part,
$params
);
print_r($response);
}
playlistItemsListByPlaylistId($service,
'snippet,contentDetails',
array('maxResults' => 25, 'playlistId' => 'id of "uploads" playlist'));
$service是Google_Service_YouTube对象。
因此,您必须从频道获取信息以检索“uploads”播放列表,该列表实际上包含该频道上传的所有视频:https://developers.google.com/youtube/v3/docs/channels/list
如果这个API是新的,我强烈建议将代码样例从默认代码片段转换为完整的样例。
因此,从一个频道检索所有视频的基本代码可以是:
class YouTube
{
const DEV_KEY = 'YOUR_DEVELOPPER_KEY';
private $client;
private $youtube;
private $lastChannel;
public function __construct()
{
$this->client = new Google_Client();
$this->client->setDeveloperKey(self::DEV_KEY);
$this->youtube = new Google_Service_YouTube($this->client);
$this->lastChannel = false;
}
public function getChannelInfoFromName($channel_name)
{
if ($this->lastChannel && $this->lastChannel['modelData']['items'][0]['snippet']['title'] == $channel_name)
{
return $this->lastChannel;
}
$this->lastChannel = $this->youtube->channels->listChannels('snippet, contentDetails, statistics', array(
'forUsername' => $channel_name,
));
return ($this->lastChannel);
}
public function getVideosFromChannelName($channel_name, $max_result = 5)
{
$this->getChannelInfoFromName($channel_name);
$params = [
'playlistId' => $this->lastChannel['modelData']['items'][0]['contentDetails']['relatedPlaylists']['uploads'],
'maxResults'=> $max_result,
];
return ($this->youtube->playlistItems->listPlaylistItems('snippet,contentDetails', $params));
}
}
$yt = new YouTube();
echo '<pre>' . print_r($yt->getVideosFromChannelName('CHANNEL_NAME'), true) . '</pre>';
下面是返回频道下所有视频id的代码
<?php
$baseUrl = 'https://www.googleapis.com/youtube/v3/';
// https://developers.google.com/youtube/v3/getting-started
$apiKey = 'API_KEY';
// If you don't know the channel ID see below
$channelId = 'CHANNEL_ID';
$params = [
'id'=> $channelId,
'part'=> 'contentDetails',
'key'=> $apiKey
];
$url = $baseUrl . 'channels?' . http_build_query($params);
$json = json_decode(file_get_contents($url), true);
$playlist = $json['items'][0]['contentDetails']['relatedPlaylists']['uploads'];
$params = [
'part'=> 'snippet',
'playlistId' => $playlist,
'maxResults'=> '50',
'key'=> $apiKey
];
$url = $baseUrl . 'playlistItems?' . http_build_query($params);
$json = json_decode(file_get_contents($url), true);
$videos = [];
foreach($json['items'] as $video)
$videos[] = $video['snippet']['resourceId']['videoId'];
while(isset($json['nextPageToken'])){
$nextUrl = $url . '&pageToken=' . $json['nextPageToken'];
$json = json_decode(file_get_contents($nextUrl), true);
foreach($json['items'] as $video)
$videos[] = $video['snippet']['resourceId']['videoId'];
}
print_r($videos);
注意:你可以在
登录https://www.youtube.com/account_advanced。
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
在最初的问题被问到之后很久才发布,但我做了一个python包,使用一个非常简单的API来做到这一点。它把所有的视频上传到一个频道,但我不确定这一部分(包括在最初的问题中):
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…
也许YouTube在这个问题发布后的8年里发生了变化,但如果没有,我制作的软件包可能不包括这个案例。
使用API:
pip3 install -U yt-videos-list # macOS
pip install -U yt-videos-list # Windows
# if that doesn't work, try
python3 -m pip install -U yt-videos-list # macOS
python -m pip install -U yt-videos-list # Windows
然后打开一个python解释器
python3 # macOS
python # Windows
然后运行程序:
from yt_videos_list import ListCreator
lc = ListCreator()
help(lc) # display API information - shows available parameters and functions
my_url = 'https://www.youtube.com/user/1veritasium'
lc.create_list_for(url=my_url)
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