Lva Gaming
@lvagaming112 • Joined December 3, 2021
At a glance
Lva Gaming ranks #7,422,446 globally for subscribers (7).
Subscribers
7
Views
0
Videos
0
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Views per subscriber
0×
Total views ÷ subscribers
Analytics tables
About
Chào mừng bạn đến với kênh YouTube của Yva Meow ! Welcome to Yva Gaming YouTube Channel ! ► Hãy nhấn ĐĂNG KÝ để là người đầu tiên xem các sản phẩm mới nhất của mình nhé ! https://bit.ly/3De90YY ▶ Theo Dõi kênh ♫ Youtube: https://bit.ly/3lwVMQV ▶ Liên hệ nếu phát hiện vấn đề liên quan đến bản quyền ! Gmail: Yvagaming…
- Handle
- @lvagaming112
- Joined
- December 3, 2021
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.