蛋哥老粉
@蛋哥老粉-r8e • Joined February 3, 2019
At a glance
蛋哥老粉 ranks #1,316,024 globally for subscribers (13.8K), #1,392,221 globally for views (2.45M), 90 videos published.
Subscribers
13,886
Views
2,457,075
Videos
90
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
27.3K
Total views ÷ total videos
Views per subscriber
177×
Total views ÷ subscribers
Subs per video
154
Subscribers ÷ total videos
Analytics tables
About
如果10訂閱………… 如果100訂閱………… 如果1000訂閱…開心 如果10000訂閱超開心 如果100000訂閱超級開心 如果1000000訂閱了超超超超超超……超超…超超超超超超…………超超超超超超……超超…超超……超級開心 目標100000訂閱 拜託大家多多訂閱我🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
- Handle
- @蛋哥老粉-r8e
- Joined
- February 3, 2019
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.