Generate a unique class for each img element randomly

Question

Generate a unique class for each img element randomly

Is there a way to assign each image a unique random class instead of giving all the images the same random class?

Any help would be appreciated.

$(document.body).ready(function () {
    bgImageTotal = 5;
    randomNumber = Math.round(Math.random() * (bgImageTotal - 1)) + 1;
    $('img').addClass('img' + randomNumber);
});

.img1{max-height:10px}
.img2{max-height:40px}
.img3{max-height:70px}
.img4{max-height:100px}
.img5{max-height:130px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

javascript jquery html css

Answer 1

Answer №1

The process of looping, generating random numbers, and assigning them:

$(document.body).ready(function () {
    var totalImages = 5;
    $('img').each(function() {
        var randomNumber = Math.round(Math.random() * (totalImages - 1)) + 1;
        $(this).addClass('img' + randomNumber);
    });
});

An important point to note is that $('img') will select all the images - hence, you need to loop through them and assign classes individually.

Answer 2

The process of looping, generating random numbers, and assigning them:

$(document.body).ready(function () {
    var totalImages = 5;
    $('img').each(function() {
        var randomNumber = Math.round(Math.random() * (totalImages - 1)) + 1;
        $(this).addClass('img' + randomNumber);
    });
});

An important point to note is that $('img') will select all the images - hence, you need to loop through them and assign classes individually.

Answer 3

Answer №2

To ensure each img element has a unique random class, you can generate an array from a set:

$(document.body).ready(function () {
    var $imgs = $('img');
    var len = $imgs.length;
    
    var set = new Set;
    while (set.size != len) {
        set.add((Math.random() * len | 0) + 1);
    }
    var arr = [...set];
    $imgs.each(function(pos) { $(this).addClass('img' + arr[pos]) });
});

.img1{height:10px}
.img2{height:40px}
.img3{height:70px}
.img4{height:100px}
.img5{height:130px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

Answer 4

To ensure each img element has a unique random class, you can generate an array from a set:

$(document.body).ready(function () {
    var $imgs = $('img');
    var len = $imgs.length;
    
    var set = new Set;
    while (set.size != len) {
        set.add((Math.random() * len | 0) + 1);
    }
    var arr = [...set];
    $imgs.each(function(pos) { $(this).addClass('img' + arr[pos]) });
});

.img1{height:10px}
.img2{height:40px}
.img3{height:70px}
.img4{height:100px}
.img5{height:130px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

Answer 5

Answer №3

The issue arises from the fact that a single random number is generated and applied to all images. To fix this, create a function within the addClass() method that will iterate over each matched element and generate a unique random number for each.

$(document.body).ready(function() {
  bgImageTotal = 5;
  $('img').addClass(function() {
    return 'img' + (Math.round(Math.random() * (bgImageTotal - 1)) + 1);
  });
});

.img1 { max-height: 10px; }
.img2 { max-height: 40px; }
.img3 { max-height: 70px; }
.img4 { max-height: 100px; }
.img5 { max-height: 130px; }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

Answer 6

The issue arises from the fact that a single random number is generated and applied to all images. To fix this, create a function within the addClass() method that will iterate over each matched element and generate a unique random number for each.

$(document.body).ready(function() {
  bgImageTotal = 5;
  $('img').addClass(function() {
    return 'img' + (Math.round(Math.random() * (bgImageTotal - 1)) + 1);
  });
});

.img1 { max-height: 10px; }
.img2 { max-height: 40px; }
.img3 { max-height: 70px; }
.img4 { max-height: 100px; }
.img5 { max-height: 130px; }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

Answer 7

Answer №4

Great question! Don't forget to incorporate a loop into your code - that's the missing piece you need. You can use a simple JavaScript for loop or utilize jQuery each if you're already working with jQuery. Give this a try and everything should function smoothly:

Your CSS:

.img1{max-height:10px}
.img2{max-height:40px}
.img3{max-height:70px}
.img4{max-height:100px}
.img5{max-height:130px}

Your HTML:

<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

And your JavaScript (jQuery):

"use strict";
  $(document).ready(function(){
    // In jQuery, you can select elements by ID, Class, Tag, etc. This is selecting all <img> tags on your page.
    $('img').each(function(i, e) {
       var imgNo = Math.floor(Math.random() * i + 1);
         $(this).addClass('img' + imgNo);
    });
  });

Everything should be up and running now, hope this solution helps!

Answer 8

Great question! Don't forget to incorporate a loop into your code - that's the missing piece you need. You can use a simple JavaScript for loop or utilize jQuery each if you're already working with jQuery. Give this a try and everything should function smoothly:

Your CSS:

.img1{max-height:10px}
.img2{max-height:40px}
.img3{max-height:70px}
.img4{max-height:100px}
.img5{max-height:130px}

Your HTML:

<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">
<img src="http://i.imgur.com/YeDYzSR.png">

And your JavaScript (jQuery):

"use strict";
  $(document).ready(function(){
    // In jQuery, you can select elements by ID, Class, Tag, etc. This is selecting all <img> tags on your page.
    $('img').each(function(i, e) {
       var imgNo = Math.floor(Math.random() * i + 1);
         $(this).addClass('img' + imgNo);
    });
  });

Everything should be up and running now, hope this solution helps!

Answer 9

Answer №5

The dollar sign function encapsulates the specified element tag within a jQuery object that includes all img elements, along with a special jQuery method that executes a function on each element in the jQuery collection by passing in the element's index and DOM element. When using .addClass on the DOM element, it needs to be enclosed within a jQuery collection. For assigning properties to individual elements, utilize the jQuery .each method.

 $('img').each(function(index, element){
   $(element).addClass('img' + randomNumber);
 })

Answer 10

The dollar sign function encapsulates the specified element tag within a jQuery object that includes all img elements, along with a special jQuery method that executes a function on each element in the jQuery collection by passing in the element's index and DOM element. When using .addClass on the DOM element, it needs to be enclosed within a jQuery collection. For assigning properties to individual elements, utilize the jQuery .each method.

 $('img').each(function(index, element){
   $(element).addClass('img' + randomNumber);
 })

Generate a unique class for each img element randomly

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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