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Is there a way to assign half of the amount of players to a certain team?
Hi, I'm trying to find a way to divide the amount of given players in half when placing them on certain teams.
This is my code:
1 | local player = game.Players:GetPlayers() |
2 | [math.random(2,#game.Players:GetPlayers())] |
3 | player.TeamColor = BrickColor.new("Bright green") |
As you can see, the obvious issue is that if the game only had 2 players, they'd both be on one team. I'm not sure how to do this, so any help is greatly appreciated.
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Maybe try checking the amount of people on both teams and if one is higher than the other, then place them on the opposite side PoePoeCannon 519 — 6y
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I'm working on the script for you, I'm almost done. Tligtre123 53 — 6y
1 answer
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1
Hey, I have made the script for you. If you have any questions, please ask me.
01 | --Get the Player count |
02 | local playercount = #game.Players:GetChildren() |
03 |
04 | --Calculate how many players should go on the team |
05 | --Is the number even? |
06 | if playercount % 2 == 0 then |
07 | --number is even |
08 | local playersselected = 0 |
09 | local toselectplayers = playercount / 2 |
10 | for i, v in pairs(game.Players:GetChildren()) do |
11 | if playersselected < toselectplayers then |
12 | v.Neutral = false |
13 | v.TeamColor = BrickColor.new("Bright green") |
14 | playersselected = playersselected + 1 |
15 | end |
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1
Thank you, this worked. Ramshackles 14 — 6y
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Glad I could help you :D Tligtre123 53 — 6y
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Just a tip, you can do local toselectplayers = math.ceil(playercount/2) instead of local toselectplayers = playercount / 2 + 0.5, this will round the number up, math.floor() will round it down. jackfrost178 242 — 6y
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Thanks! I didn't know that! Tligtre123 53 — 6y