I have the below html:
<div id="one">
<div class="two">
<div开发者_如何学Python class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
</div>
</div>
in jQuery if I am to wrap every 3 elemnts in a new div like below how could I do that?
<div id="one">
<div class="two">
<div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
</div>
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
</div>
<div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
<div class="entry">
Content here
</div>
</div>
</div>
</div>
I tried slice and wrap which did not really help me. Any ideas? appreciate your help on this.
Thanks, L
var entries = $('#one > .two > div.entry');
entries.each(function(i) {
if( i % 3 == 0 ) entries.slice(i,i+3).wrapAll('<div>');
});
http://jsfiddle.net/PZkSL/
Refer to this How to wrap every 3 child divs with html using jquery? or Wrap every 3 divs in a div
How about getting the div element with class "two", and iterate through it's children. When you find the first child use $(firstChild).before('<div>
'); and on the third child use $(thirdChild).after('</div>
'); Just iterate all the way through using this method. Haven't used .before() or .after() so not sure how this will pan out. Good luck.
I would just write a simple for
loop, something like:
var two = $(".two");
var elements = $(".entry");
var group = [];
for(var i=0;i<elements.length;i++) {
var mod = i % 3;
if(mod == 0 && group.length > 0) {
var container = $('<div></div>');
two.append(container);
for(var j=0;j<group.length;j++) {
container[0].appendChild(group[j]);
}
group = [];
}
group[mod] = elements[i];
}
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