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I use uBlock Origin and PrivacyBadger (which are both free). Does Adguard work better/differently to justify the cost?
by Spare_account 6y ago
I use uBlock Origin and PrivacyBadger (which are both free). Does Adguard work better/differently to justify the cost?
- ffpip 6y agoDon't use both. Use uBlock Origin. Never, ever use 2 ad blockers in a single layer (here, browser). Privacy Badger also changed how it works. Making it just another adblocker. So stick with uBO
- h2onock 6y agoI’m curious, why should you not use 2? I use Privacy Badger and ublock and haven’t had any problems that I know of.
- jdeibele 6y agoNot original poster but I recommend Adguard for your DNS: https://adguard.com/en/adguard-dns/overview.html https://adguard.com/en/adguard-dns/overview.html I changed the router settings for the house six months ago and have had no complaints. We've got 3 teenagers doing online high school or university and no issues. Once in a great while I get a message from a site that says "we've detected you're using an adblocker" but really no issues. There's no control over what is blocked or not blocked. That worried me but has turned out to work just fine. There are other solutions that do allow you to maintain a list but charge an annual fee. https://duckduckgo.com/?t=ffab&q=adblocking+dns https://duckduckgo.com/?t=ffab&q=adblocking+dns shows a lot more options than there were six months ago.
- petecooper 6y ago>Does Adguard work better/differently to justify the cost? I can't comment on better as I haven't used either of those products in anger or recently, but Adguard installs itself as a system app, not a browser plugin, and it just works. It's become a standard install on new computers when I set them up. The only side effect I see of an average day is some pages taking a second or two longer before the first paint, presumably due to the quantity of ads removed. For where I am, the cost of an evergreen licence for a single seat is less than an hour of work, it looks after itself, and I only see ads in Tor. Well worth the $.