壞壞
@壞m • Joined December 25, 2024
At a glance
壞壞 ranks #1,782,929 globally for subscribers (5.9K), #1,869,876 globally for views (1.03M), 173 videos published from TW.
Subscribers
5,900
Views
1,038,000
Videos
173
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
6K
Total views ÷ total videos
Views per subscriber
176×
Total views ÷ subscribers
Subs per video
34
Subscribers ÷ total videos
Analytics tables
About
訂閱按讚 感謝支持！ 平常喜歡玩射擊遊戲.即時戰略.MMO手遊 也是老遊戲的愛好者 秉持著不服就上的精神 開幹！ 天堂遊戲群：https://reurl.cc/8baxGj 壞孩子(FB)：https://reurl.cc/RLm66G 聯絡壞孩子：[email protected] 聯絡壞孩子DC：_369369369(前面有底線)
- Handle
- @壞m
- Joined
- December 25, 2024
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.