CODEWITHSUNDEEP
javascriptjqueryexpressionengine
cusejuice
cusejuice

Reputation: 10691

Insert <div> for every 5 elements using Javascript

I have a simple list of images that is being controlled via a CMS (ExpressionEngine). Like this:

<div class="wrapper">
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
    <a href="#"><img src="#" /></a>
</div>

What I want to do is for every 5 images, wrap them in a div with a class of "slide." To look like this:

<div class="wrapper">
    <div class="slide">
      <a href="#"><img src="#" /></a>
      <a href="#"><img src="#" /></a>
      <a href="#"><img src="#" /></a>
      <a href="#"><img src="#" /></a>
      <a href="#"><img src="#" /></a>
    </div>
    <div class="slide">
      <a href="#"><img src="#" /></a>
      <a href="#"><img src="#" /></a>
      <a href="#"><img src="#" /></a>
    </div>
</div>

The reason I am not manually coding the "" in is because of a jQuery content slider that I am using which requires every 5 images to be wrapped inside a slide div.

I'm not sure how what the code in ExpressionEngine would be to do this, but I figure it might just be easier to use Javascript to wrap every 5 images with the div. And to just have ExpressionEngine output the different images all at once.

Any help?

Upvotes: 9

Views: 12117

Answers (6)

Guffa
Guffa

Reputation: 700432

You can just create a div for every fith element and move the links into them using the append method:

var wrapper = $('.wrapper');
var div;
$('a', wrapper).each(function(i,e){
    if (i % 5 == 0) div = $('<div/>').addClass('slide').appendTo(wrapper);
    div.append(e);
});

Demo: http://jsfiddle.net/Guffa/ybrxu/

Upvotes: 4

bcoughlan
bcoughlan

Reputation: 26627

Use slice() to select the element subset then wrapAll() to wrap the div around them. Here's a function that does that.

var wrapEveryN = function(n, elements, wrapper) {
   for (var i=0; i< elements.length; i+=n) {
       elements.slice(i,i+n).wrapAll(wrapper);
   }
}

wrapEveryN( 5, $(".wrapper a"), '<div class="slide"></div>' );

Demo: http://jsfiddle.net/C5cHC/

Note that the second parameter of slice may go out of bounds, but jQuery seems to handle this automatically.

Upvotes: 0

Chandu
Chandu

Reputation: 82923

Try this:

$(function(){
    var curDiv = null;
    var mainDiv = $("div.wrapper");
    $("span", mainDiv).each(function(i, b){
        if(i%5 == 0) {
            curDiv = $("<div class='slide'/>").appendTo(mainDiv);
        }
        curDiv.append(b);
    });
});

Upvotes: 1

Mark Schultheiss
Mark Schultheiss

Reputation: 34176

I think this would do that:

var links = $('.wrapper').children();
for (var i = 0, len = links.length; i < len; i += 5) {
    links.slice(i, i + 5).wrapAll('<div class="slide"/>');
}

Upvotes: 1

user113716
user113716

Reputation: 322502

Here's one way:

Example: http://jsfiddle.net/T6tu4/

$('div.wrapper > a').each(function(i) {
    if( i % 5 == 0 ) {
        $(this).nextAll().andSelf().slice(0,5).wrapAll('<div class="slide"></div>');
    }
});

Here's another way:

Example: http://jsfiddle.net/T6tu4/1/

var a = $('div.wrapper > a');

for( var i = 0; i < a.length; i+=5 ) {
    a.slice(i, i+5).wrapAll('<div class="slide"></div>');
}

Upvotes: 25

Ahmad Alfy
Ahmad Alfy

Reputation: 13371

You need to use jQuery slice with wrap

Check this question

Upvotes: 0

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