'how to load background images into different blocks from an array
I have 9 blocks, where it is necessary that when reloading the background image, pictures from the folder are loaded without repeating. How can I implement this?
When I click on the button, I reload the page, then I interfere with the array. How then to get these pictures?
<div class="game">
<div></div>
<div></div>
<div></div>
<div></div>
<div></div>
<div></div>
<div></div>
<div></div>
<div></div>
</div>
const imgCollection = [
"assets/img/1.png",
"assets/img/2.png",
"assets/img/3.png",
"assets/img/4.png",
"assets/img/5.png",
"assets/img/6.png",
"assets/img/7.png",
"assets/img/8.png",
"assets/img/9.png",
];
reset.addEventListener("click", () => {
location.reload();
x=imgCollection.sort(()=>Math.random()-0.5);
});
Solution 1:[1]
The styling is up to you, but this setup allows you to refresh the board without completely reloading the page.
HTML:
<div class="game">
</div>
<button id="reload-btn">reload</button>
Javascript:
let gameEl = document.querySelector('.game');
let reloadBtn = document.querySelector('#reload-btn');
const imgCollection = [
"assets/img/1.png",
"assets/img/2.png",
"assets/img/3.png",
"assets/img/4.png",
"assets/img/5.png",
"assets/img/6.png",
"assets/img/7.png",
"assets/img/8.png",
"assets/img/9.png",
];
const shuffle = (array) => {
for (let i = array.length - 1; i > 0; i--) {
const j = Math.floor(Math.random() * (i + 1));
[array[i], array[j]] = [array[j], array[i]];
}
return array;
}
const drawBlocks = (imageArray) => {
gameEl.innerHTML = "";
shuffle(imageArray).forEach((image) => {
let newBlock = document.createElement('div');
newBlock.innerHTML = image;
newBlock.style.backgroundImage = `url('${image}')`;
gameEl.appendChild(newBlock);
})
}
reloadBtn.addEventListener("click", () => {
drawBlocks(imgCollection);
});
live example: https://codepen.io/jriches/pen/OJOepNP
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Jacob Riches |
