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A sus command with percentage statements

Asked Oct 15, 2020·4 replies·Last activity 5 years ago·Open in Discord ↗
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Mauve LlamaOriginal post

Hey All!

Hope you're well?

I'm trying to create a conditional command for a !sus.... i am struggling to piece it all together.
The command i'm hoping to make is

Random Number as a variable.... then if Number is below 25 = $user is 'Number'% sus.... I think they're lying, else if number is 25-50 =$user is 'Number'% sus.... they seem too dodgy lets vote them, else if number is 50-75 = $user is 'Number'% sus... $user is highly suspcious.... lets get rid else if number is greater than 75 = $user is 'number'% sus.... they definitely did it!

Thanks

4 replies

Quirky Fox

Use this:

$(eval R=Math.floor(Math.random()*100+1);R<=25?`$(user) is ${R}% sus... FIRST_RESPONSE`:R>25&&R<=50?`$(user) is ${R}% sus... SECOND_RESPONSE`:R>50&&R<=75?`$(user) is ${R}% sus... THIRD_RESPONSE`:R<75?`$(user) is ${R}% sus... FOURTH_RESPONSE`:` `)

Fill in the response the first is the least sus the fourth is the most sus.

Mauve Llama

Ooooo,

Thank you it looks awesome... the only thing is that when i run the command. i get "unexpected end of input"

Quirky Fox

Sorry I forgot you have to have an outcome if none of them are true I edited the eval above it should work now.

Mauve Llama
Quirky Fox said:
```
$(eval R=Math.floor(Math.random()*100+1);R<=25?$(user) is ${R}% sus... FIRST_RESPONSE:R>25&&R<=50?$(user) is ${R}% sus... SECOND_RESPONSE:R>50&&R<=75?$(user) is ${R}% sus... THIRD_RESPONSE:R<75?$(user) is ${R}% sus... FOURTH_RESPONSE: )
```

That works perfectly now ♥ thank you so much xox