setTimeout in a loop execute all at the same time

Question

I want to open a link every 3 seconds. I am using the setTimeout function, but it doesn't work. All links will be opened once.

for(var i=0; i < url.length-1; i++) {
  setTimeout(function(){
    linkaddress=url[i];
    window.open(linkaddress);
  }, 3000);
}

1. This creates 3 "timeouts" at once, 2. `i` will be `url.length` when the callbacks from the "timeouts" is executed — Andreas, Apr 18 '19 at 06:59

jacqbus · Accepted Answer · 2019-04-18T07:19:09.057

Use "let" instead of "var" for block level scoping, and then multiply your time by i variable (more info). Code:

 var url = ["https://domain1.com","https://www.domain2.com"],
     timeout = 3; // Time in second

 for(let i=1; i <= url.length; i++){
   setTimeout(function(){
     linkaddress=url[i-1];
     window.open(linkaddress);
   }, i * timeout * 1000);
 }

EDIT: Note that this code uses EcmaScript 6 features

score 1 · Answer 2 · answered Apr 18 '19 at 07:15

Use setInterval instead

url = ['a', 'b', 'c'];

var i = 0;
var interval = setInterval(function() {
  if (i <= url.length - 1) {
    ///linkaddress = url[i];
    //window.open(linkaddress);
    console.log(url[i]);
    i++;
  } else {
    clearInterval(interval);
  }
}, 3000);

setTimeout in a loop execute all at the same time

2 Answers2

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