我该如何获得孩子们的前四个并将其包装到div中? [英] How can I get first 4 of the childrens and wrap it into the div?

问题描述

如何找到父级的前4个子级并将其包装到div中，然后再封装4个，以此类推?谷歌不知道！ :(

How can I find first 4 children of the parent and wrap it into div, then next 4 and so on? Google does not know! :(

从:

<div class="main">
    <div>1</div>
    <div>2</div>
    <div>3</div>
    <div>4</div>
    <div>5</div>
    <div>6</div>
    <div>7</div>
    <div>8</div>
</div>

收件人:

<div lass="main">
    <div class="pack1">
        <div>1</div>
        <div>2</div>
        <div>3</div>
        <div>4</div>
    </div>

    <div class="pack1">
        <div>5</div>
        <div>6</div>
        <div>7</div>
        <div>8</div>
    </div>
</div>

任何帮助，不胜感激.

Any help much appreciated.

我可以通过以下方式找到前4个:$('.main > div:lt(4)').wrapAll('<div class="pack1"></div>'); 但是如何获得每四个??

I can find first 4 by: $('.main > div:lt(4)').wrapAll('<div class="pack1"></div>'); But how to get every 4??

Pete

推荐答案

不是最优雅的方法，但是它的工作原理.

Not the most elegant way but its working.

var sel;
while ( (sel = $('.main > div:not(.pack1)')).length > 0 )
{
    sel.slice(0,4).wrapAll('<div class="pack1"></div>');
}

这里是小提琴.

在这里有该课程的柜台.我们必须使用第二类pack来排除所有打包的div.

Here with counter for the class. We have to use a 2nd class pack for the exclude of all packed divs.

var sel;
var count = 1;
while ( (sel = $('.main > div:not(.pack)')).length > 0 )
{
    sel.slice(0,4).wrapAll('<div class="pack pack'+ count++ +'"></div>');
}

和小提琴

或者我们一起去

var sel;
var count = 1;
while ( (sel = $('.main > div:not([class^="pack"])')).length > 0 )
{
    sel.slice(0,4).wrapAll('<div class="pack'+ count++ +'"></div>');
}

为此，

小提琴.

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