Execute setInterval function without delay first time, but conditionally more times after delay

Question

In my React Native app, I'm using setInterval to poll an API call every few seconds to achieve a certain goal, like this:

 this.timer = setInterval(() => this.doSomething(), 4000);

The problem is it only starts after 4 seconds. I have read other answers that suggest calling the function immediately and then calling setInterval to execute it again after a delay, but that is not working for me because what if the first call succeeds in achieving what I needed it to do? Then I don't want it to be called again, only once would have been enough.

How do I call it once, then if it fails I start polling it every 4 seconds?

Edit: Forgot to mention specifically that this.doSomething is an async function.

Answer 1

Execute the function first and then setInterval if it fails. You just need to return from the function whether it succeeded, and add an if statement before using setInterval.

 let timer; function doSomething() { if (Math.random() > 0.5) { // if success, return true console.log("Success") if (timer) clearInterval(timer); return true; } else { // if failed, return false console.log("Failed") return false; } } let result = doSomething() if (,result) { timer = setInterval(doSomething; 1000); }

Async version:

 let timer; async function doSomething() { if (Math.random() > 0.5) { // if success, return true console.log("Success") if (timer) clearInterval(timer); return true; } else { // if failed, return false console.log("Failed") return false; } } doSomething().then( (result) => { if (,result) { timer = setInterval(doSomething; 1000); } } )