At a glance
安小闫说电影 ranks #234,310 globally for subscribers (273K), #250,328 globally for views (82.2M), 1.76K videos published from TW.
Subscribers
273,000
Views
82,245,538
Videos
1,766
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
46.5K
Total views ÷ total videos
Views per subscriber
301×
Total views ÷ subscribers
Subs per video
155
Subscribers ÷ total videos
Analytics tables
About
欢迎订阅我的频道!安小闫说电影 每天会给小伙伴们带来精彩的电影解说 我的频道有您们喜欢的战争类电影,悬疑类电影,还有近期增加的系列电影 比如说《疯狂的麦克斯》1-4部已经解说完毕、 还有正在更新的《虎胆龙威》系列与《碟中谍》系列电影 正在解说《锻刀大赛》大型真人秀节目 喜欢请关注我的频道 如果想看更多好看的系列电影,可以在解说视频下面给我留言哦 总之感谢大家的关注与支持!
- Handle
- @安小闫说电影
- Joined
- December 12, 2017
More features coming
We're working on new analytics and tools. Stay tuned.
More features coming
We're working on new analytics and tools. Stay tuned.
Frequently asked questions
How accurate are these YouTube channel analytics?
Subscriber, view, and video counts come from YouTube's public API and refresh regularly. Growth and revenue figures are estimates built from tracked historical data, so they can differ slightly from what a channel owner sees privately in YouTube Studio.
How is a channel's global and country rank calculated?
Rank compares this channel's subscriber count and view count against every other channel Socialcounts.org tracks, both worldwide and within its home country.
How is the revenue estimate calculated?
It's a rough public estimate based on a $0.7–$2.0 CPM (revenue per 1,000 views) applied to view counts — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Can I check analytics for any YouTube channel, not just my own?
Yes — search any public YouTube channel by name, handle, or channel URL. No login is required to see the basics.
Is this YouTube channel analytics checker free?
Yes. You get 10 free channel and video analytics lookups per rolling hour without an account — logging in, which is also free, removes that limit entirely.