如何在 For 循环中使用超时？

Question

So, I'm trying to use create a loop which does this:

setTimeout(function() {           
    console.log("Hey!");

    setTimeout(function() {           
       console.log("Hey!");

       setTimeout(function() {           
       console.log("Hey!");

       }, 1000);

    }, 1000);

 }, 1000);

So, I tried it like this.

for (i = 0; 1 < 3; i++){
    setTimeout(function() {           
    console.log("Hey!");
}, 1000);
}

How ever, it's not working.

Doing some research I've noticed this is because the timeOuts are getting added to each other with each loop. How can I work around this?

Answer 1

You have to recursively call the time outs, so write a function that takes an argument of the current number of attempts. Have it do an operation, and then call itself with the attempts argument += 1.

You should pass a number of attempts as a safeguard so you can tell the function not to call itself if the attempts number is > some limit, to avoid infinite loops.

Something like:

timedLog(attempts) {
  console.log('Hey!');
  if (attempts > 10) {
    return;
  } else {
    setTimeout(function() { timedLog(attempts + 1); }, 1000);
  }
}

It doesn't look like a for loop anymore, but it's the same principle.

Answer 2

If the output of each function is the same, and you want to print it in the same interval, then you could use setInterval .

function myInterval() {
  return setInterval(function(){
    console.log("Hey!");
  }, 1000);
};

var id = myInterval();

It will repeat forever, so you would have to stop it, in this case, after 3000 ms.

setTimeout(function(){
   clearInterval(id);
}, 3000);

Answer 3

Just use a loop, increasing the timeout interval each time:

for (let i = 0; 1 < 3; i++){
    setTimeout(function() {           
        console.log("Hey!");
    }, i*1000);
}

(Note, that if your callback function depended on i , this won't work as expected if you use var instead of let in the for loop header. The reason is to do with closures - there are many questions about this on SO. But it works perfectly with let - and there are other simple enough fixes if for some reason you can't use let .)