const divArr = [];
for (let i = 0; i < 5; i++) {
document.getElementById("h").innerHTML += `<div id=${i}>hello</div>`;
divArr.push(document.getElementById(`${i}`));
}
console.log(divArr);
let divArrIndex = 0;
setInterval(() => {
document.getElementById(`${divArrIndex}`).style.backgroundColor = 'green';
if (divArrIndex > 4) {
divArrIndex = 0;
}
divArrIndex += 1;
}, 1000 / 1);<div id="h">alsdjf;lkajsdf</div>But, when I use
divArr[divArrIndex].style.backgroundColor = "green"
instead of
document.getElementById(`${divArrIndex}`).style.backgroundColor='green';
I only get the last div green. Why? codepen: https://codepen.io/sai-nallani/pen/KKopOXZ?editors=1111
By reassignment to innerHTML, you are destroying and recreating the contents of #h in each iteration of the loop. You create #0, then discard it and create #0 and #1, then discard those and create #0, #1, #2... So the elements you push into the array don't exist any more in your document (though references to them in divArr still keep the garbage collector from destroying them outright).
When you change the colour of divArr[0], you are changing the colour of an element that only exists in memory, but is not in DOM any more. However, #4 is still the original #4, it has not been discarded, since you have performed no further assignments to innerHTML.
One solution is to gather all the divs after you have constructed them all. You can use another loop, but the easiest way would be:
const divArr = Array.from(document.querySelectorAll('#h > div'));
(Depending on what you are doing with it next, you may not need Array.from since NodeList should suffice for many purposes.)
Another solution is to construct elements in their own right, not by changing the parent's innerHTML:
const hEl = document.querySelector('#h');
for (let i = 0; i < 5; i++) {
const divEl = document.createElement('div');
divEl.textContent = 'Hello';
divEl.id = i;
hEl.appendChild(divEl);
divArr.push(divEl);
}
This way, every element is created and added to #h without affecting any other elements already there.