dan @daanmoura.bsky.social · 18/02/2025Eu preciso de uma IA para organizar o meu Watch later do YouTube 010
Reposted by dan🚨 AI News | TestingCatalog @testingcatalog.com · 24/01/2025BREAKING 🚨: OpenAI released Canvas rendering for HTML & React and now, Canvas feature is available with o1 model as well 🔥 The new rendering feature is now available to Free users as well 👀 081
dan @daanmoura.bsky.social · 22/01/2025queria tirar uma foto (de verdade, sem IA), um auto retrato, igual a que tenho no meu perfil. Mas ainda não acertei o corte de cabelo e iluminação. 000
Reposted by danWes Bos @wesbos.com · 21/01/2025You should be using JavaScript sets more often Now in all browsers and Node - these 7 new set methods are key for when trying to compare two arrays or sets of data. 511600148
dan @daanmoura.bsky.social · 17/01/202521st.dev this is gold ✨ #bolhadev @sseraphini.bsky.social21st.dev21st.dev - The NPM for Design EngineersShip polished UIs faster with ready-to-use React Tailwind components inspired by shadcn/ui. Built by design engineers, for design engineers. 053
Reposted by danDaniel #Ofertas Nintendo Reenlsober @danielreen.bsky.social · 16/10/2024#SORTEIO 🎉 A @Nuuvem liberou a chave de Super Mario Party Jamboree pra vocês! Só seguir os passos com atenção: - Me segue - 💙Curtir o post +🔄 RT sem comentar - Preencha o formulário bit.ly/sorteio-jamb... e comente aqui quais jogos de Mario Party jogou! 854978870
dan @daanmoura.bsky.social · 17/09/2024Bluesky now has over 10 million users, and I was #723,584! 120
![New JavaScript Set Methods
const set1 = new Set(['wes', 'kait']);
const set2 = new Set(['wes', 'scott']);
Difference
Elements in set2 but not in set1
set2.difference(set1)
> ['scott']
Intersection
Elements that exist in both sets
set1.intersection(set2)
> ['wes']
Symmetric Difference
Elements in either set, but not both
set1.symmetricDifference(set2)
> ['kait', 'scott']
Union
All elements from both sets
set1.union(set2)
> ['wes', 'kait', 'scott']
Is Disjoint From
True if sets share no elements
set1.isDisjointFrom(set2)
> false
Is Subset Of
True if all elements are in other set
set1.isSubsetOf(set2)
> false
Is Superset Of
True if it contains all other's elements
set1.isSupersetOf(set2)
> false](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:etdjdgnly5tz5l5xdd4jq76d/bafkreifpciv7ufsmynaqjmiw63rmyybije242ep2ylqs3iarkehaz4qxzi)
