At a glance
LJFS Gaming ranks #1,391,850 globally for subscribers (12.1K), #1,610,121 globally for views (1.61M), 737 videos published.
Subscribers
12,100
Views
1,617,516
Videos
737
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
2.19K
Total views ÷ total videos
Views per subscriber
134×
Total views ÷ subscribers
Subs per video
16
Subscribers ÷ total videos
Analytics tables
About
✨ROAD TO 15K🔥 Always Enjoyed by playing games and here to entertain you as well! 🖖SUBSCRIBE TO BCM OUR FAM!🔥 • Channel Achievements 🔥 500🏆-6 JUN 2023 1K🏆-4 AUG 2024 5K🏆-21 MAR 2025 10K🏆-28 APR 2025
- Handle
- @sj_og7
- Joined
- June 9, 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.