ASMR for You
@asmr-asleep • Joined May 7, 2026
At a glance
ASMR for You ranks #6,172,400 globally for subscribers (30), #5,127,011 globally for views (9.70K), 237 videos published from TW.
Subscribers
30
Views
9,709
Videos
237
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
41
Total views ÷ total videos
Views per subscriber
324×
Total views ÷ subscribers
Subs per video
0
Subscribers ÷ total videos
Analytics tables
About
「我們用降低了一半音量的音樂，和你一起度過時光。」 在這個喧囂的世界裡，ASMR for You 致力於為您提供最純粹的舒壓與陪伴。我們精心調製低分貝的療癒旋律，結合細膩的環境音，陪伴您度過每一個需要助眠與療癒的時刻。無論是深夜的自我對話，還是午後的專注學習，讓我們的音樂成為您心靈最溫暖的避風港。 🌿✨
- Handle
- @asmr-asleep
- Joined
- May 7, 2026
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.