Paying to have a server = less random servers

Opmerkingen

13 opmerkingen

  • ||Jojo||
    Who cares about the dead servers?
    5
  • TheFutureKnight
    No, just no, this would defeat the whole idea of Discord, gathering gamers and being free.
    0
  • Gogetters

    Hell no, big servers are cancer.

    1
  • Rexowogamer
    NO GOD PLEASE NO... The whole point of Discord is so people can freely and easily chat with friends or strangers to play games, it’s MEANT to be free. Who gives a f*ck about dead servers? Just leave or ignore them!
    1
  • Costpap
    Umm... No one ever asked for this? Like, there's literally no reason to do it. It will do more damage than what it's meant to solve.
    0
  • index.ts
    just dont join "random servers"...Discord is a *free platform*. If you want to pay for a server, use teamspeak. The fact that servers are free is the main reason that people use Discord
    0
  • undermaster.
    Discord is supposed to be a free platform. Paying to own a server is so useless, Discord is for making friends and we need a server for it, this is like paying to have a friend
    0
  • pepsiplayzz
    Not a good idea, as the soul purpose of Discord is to be free and bring together gamers in servers. You don't need to join random "dead" servers, just leave them.
    1
  • nabs

    no way, it is not worth it

    0
  • Exaint
    Big -1, having free servers is 1 of the core points of discord. And this fundamental feature should always be free
    0
  • Karate Wumpus
    -1, The fun part about discord is having a lot of random fun servers that get used in some way. Thats why we use discord
    0
  • luigi25

    Why does everything have to be about money? This is like paying to have friends! This is too much like real life! OMG! If I'm having to pay to be on discord, I might as well not even bother. MIght as well go back to Facebook. I'd get farther with that!

    1
  • felix

    N O 

    just no

    discord is made a FREE chat

    they get money with nitro

    and dont join if dont like

    0

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