At a glance
Zina ranks #2,031,641 globally for subscribers (3.73K), #2,254,782 globally for views (518K), 109 videos published.
Subscribers
3,730
Views
518,694
Videos
109
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
4.75K
Total views ÷ total videos
Views per subscriber
139×
Total views ÷ subscribers
Subs per video
34
Subscribers ÷ total videos
Analytics tables
About
嗨嗨!!大家 目前是個剛起步的小賣家 這裡是在記錄我的生活、賣家日常 新開幕的小店喲😚 ‼️小店🈵️100才出貨喔‼️ 買就送贈品 買越多贈越多(*≧ω≦) 賣場剛開始經營 所以有各種問題請多多包涵🥺 本人還是學生出貨可能比較慢 也請多多見諒囉😉 𝑊𝑒𝑏𝑠𝑖𝑡𝑒🔗 https://shopee.tw/jen0820226?smtt=0.204547637-1650472051.9 蝦皮:@jen0820226
- Handle
- @zina5199
- Joined
- June 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.