有没有办法使用gdata-python-client在youtube上搜索播放列表?至于文档是不可能的,但可能有一些解决方法......
答案 0 :(得分:1)
you tube python API似乎有一种搜索匹配特定术语的播放列表的方法。 根据文档,API可以检索与用户指定的搜索词匹配的播放列表列表。
[编辑:卷曲输出]
我刚使用curl,API似乎工作正常。
~ $ curl 'http://gdata.youtube.com/feeds/api/playlists/snippets?q=soccer&start-index=11&max-results=10&v=2'
输出:
<?xml version='1.0' encoding='UTF-8'?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearch/1.1/' xmlns:gd='http://schemas.google.com/g/2005' gd:etag='W/"C0cHRX85eCp7ImA9Wx5VEko."'><id>tag:youtube.com,2008:playlists:snippets</id><updated>2010-10-05T09:30:34.120Z</updated><category scheme='http://schemas.google.com/g/2005#kind' term='http://gdata.youtube.com/schemas/2007#playlistLink'/><title>YouTube Playlists matching query: soccer</title><logo>http://www.youtube.com/img/pic_youtubelogo_123x63.gif</logo><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://gdata.youtube.com/feeds/api/playlists/snippets?v=2'/><link rel='http://schemas.google.com/g/2005#batch' type='application/atom+xml' href='http://gdata.youtube.com/feeds/api/playlists/snippets/batch?v=2'/><link rel='self' type='application/atom+xml' href='http://gdata.youtube.com/feeds/api/playlists/snippets?q=soccer&start-index=11&max-results=10&v=2'/><link rel='service' type='application/atomsvc+xml' href='http://gdata.youtube.com/feeds/api/playlists/snippets?alt=atom-service&v=2'/><link rel='previous' type='application/atom+xml' href='http://gdata.youtube.com/feeds/api/playlists/snippets?q=soccer&start-index=1&max-results=10&v=2'/><link rel='next' type='application/atom+xml' href='http://gdata.youtube.com/feeds/api/playlists/snippets?q=soccer&start-index=21&max-results=10&v=2'/><author><name>YouTube</name><uri>http://www.youtube.com/</uri></author><generator version='2.0' uri='http://gdata.youtube.com/'>YouTube data API</generator><openSearch:totalResults>1241773</openSearch:totalResults><openSearch:startIndex>11</openSearch:startIndex><openSearch:itemsPerPage>10</openSearch:itemsPerPage></feed>~ $