Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

jQuery Loop through each div

Tags:

jquery

loops

I'm pretty sure this will be a really easy answer for you jQuery whizzes, and I'm also pretty such it involves a loop of some kind.

I'm trying to perform essentially the same calculation on two separate divs, but assigning a different CSS width value to each id based on the number of images found. The calculations I'm performing are irrelevant to my problem really, but I put them in anyway because it's the actual code I'm working with.

Here is the markup...

<div id ='test1' class='target'>
  <div class='scrolling'>
    <img/>
    <img/>
    <img/>
  </div>
</div>

<div id ='test2' class='target'>
  <div class='scrolling'>
    <img/>
    <img/>
    <img/>
  </div>
</div>

Below is my current jQuery, which works fine, but it's inefficient because I have to write another chunk of code for every div added. How can I standardise this so that it runs through every div with the class of target? Thanks

/* Measure the width of each image. */
test1 = $('#div1 .scrolling img').width();
test2 = $('#div2 .scrolling img').width();

/* Find out how many images there are. */
test1img = $('#div1 .scrolling img').length;
test2img = $('#div2 .scrolling img').length;

/* Do the maths. */
final1 = (test1 * test1img)*1.2;
final2 = (test2 * test2img)*1.2;

/* Apply the maths to the CSS. */
$('#div1 .scrolling').width(final1);
$('#div2 .scrolling').width(final2);    
like image 801
SparrwHawk Avatar asked Jan 02 '12 13:01

SparrwHawk


People also ask

How do I iterate through a div in jQuery?

jQuery Selector can be used to find (select) HTML elements from the DOM. Once an element is selected, the jQuery children() method is called to find all the child elements of the selected element.

How do I iterate through jQuery?

each(), which is used to iterate, exclusively, over a jQuery object. The $. each() function can be used to iterate over any collection, whether it is an object or an array. In the case of an array, the callback is passed an array index and a corresponding array value each time.

How do you loop through an element with the same class in jQuery?

Answer: Use the jQuery each() Method You can simply use the jQuery each() method to loop through elements with the same class and perform some action based on the specific condition.

How each loop works in jQuery?

each() jQuery's each() function is used to loop through each element of the target jQuery object — an object that contains one or more DOM elements, and exposes all jQuery functions. It's very useful for multi-element DOM manipulation, as well as iterating over arbitrary arrays and object properties.


2 Answers

Like this:

$(".target").each(function(){
    var images = $(this).find(".scrolling img");
    var width = images.width();
    var imgLength = images.length;
    $(this).find(".scrolling").width( width * imgLength * 1.2 );
});

The $(this) refers to the current .target which will be looped through. Within this .target I'm looking for the .scrolling img and get the width. And then keep on going...

Images with different widths

If you want to calculate the width of all images (when they have different widths) you can do it like this:

// Get the total width of a collection.
$.fn.getTotalWidth = function(){
    var width = 0;
    this.each(function(){
        width += $(this).width();
    });
    return width;
}

$(".target").each(function(){
    var images = $(this).find(".scrolling img");
    var width = images.getTotalWidth();
    $(this).find(".scrolling").width( width * 1.2 );
});
like image 67
Niels Avatar answered Sep 21 '22 08:09

Niels


You're right that it involves a loop, but this is, at least, made simple by use of the each() method:

$('.target').each(
    function(){
        // iterate through each of the `.target` elements, and do stuff in here
        // `this` and `$(this)` refer to the current `.target` element
        var images = $(this).find('img'),
            imageWidth = images.width(); // returns the width of the _first_ image
            numImages = images.length;
        $(this).css('width', (imageWidth*numImages));

    });

References:

  • css().
  • each().
  • find().
like image 38
David Thomas Avatar answered Sep 23 '22 08:09

David Thomas