looping a function with setTimeout

Question

for (i=0;i<5;i++){
setTimeout(bye,2000)}

console.log("asda")

function bye(){
  console.log("gudbye")
}

I want the program to log bye once evey 2 second,but this works just for the fist log, so i get instantly 5 time "gudbye" in 2 seconds

`setTimeout()` doesn't wait, it just schedules the function to run in 2 seconds then returns immediately. — Barmar, Nov 22 '20 at 02:00

Otacílio Maia · Answer 1 · 2020-11-22T02:13:46.603

0

Change your code to:

setInterval(bye,2000)

function bye(){
  console.log("gudbye")
}

setInterval will run your function every 2 seconds, instead of running just one time after 2000 seconds per call.

but if you only want it to be called 5 times you can try the recursive aproach:

bye(1)
function bye(callCount){
  console.log("gudbye")
  if(callCount < 5){
      setTimeout(() => bye(callCount+1), 2000);
  }
}

edited Nov 22 '20 at 02:13

answered Nov 22 '20 at 02:05

Otacílio Maia

But it also never stops. Whereas original poster wanted to run 5 times. – Marc Nov 22 '20 at 02:08
I'm not sure if the original poster only wants 5 times, but if its the case, it can be changes to something like: bye(0) function bye(callCount){ console.log("gudbye") if(callCount < 5){ setTimeout(() => bye(callCount+1), 2000); } } – Otacílio Maia Nov 22 '20 at 02:12

1 Answers1