jquery remove element if it occurs twice

问题

I have something like this in my html:

<div class="onlyContent">
    <!-- some more stuff here -->
</div>
<div class="onlyContent">
    <!-- some more stuff here -->
</div>

Now, with jQuery I want to remove the 2nd occurence of the class onlyContent HTML from the dom.

The final result should be this:

<div class="onlyContent">
    <!-- some more stuff here -->
</div>

I figured out that you can somehow use the nth-child-selector, but trying to access it this way didn't do it for me

$('.onlyContent:nth-child(2)').remove();

回答1:

You can use :eq(1) for targetting second element in matched set:

$('.onlyContent:eq(1)').remove();

If the number of elements more than two, then you should use :not(:first) or :gt(0):

$('.onlyContent:not(:first)').remove();

$('.onlyContent:gt(0)').remove();

回答2:

use below code . use jQuery :eq() selector.

Select the element at index n within the matched set.

check DEMO

   $('.onlyContent:eq(1)').remove();

回答3:

you can try this

$('.onlyContent:gt(0)').remove();

回答4:

You can try this -

   $('.onlyContent:gt(0)').remove();

It will remove all the duplicates. Only the first one will be present.

回答5:

You can use .slice for this.

$(".onlyContent").slice(1).remove();

Nice and simple, no fuss. Working example.

From the .slice documentation:

Reduce the set of matched elements to a subset specified by a range of indices

You can find more here.

回答6:

Use .length property to check the numbers of selected elements on the page and if its value is greater than 1 then remove the elements other than first...

if($('.onlyContent').length > 1) {
$('.onlyContent:gt(0)').remove();
}

回答7:

you can try

$(".onlyContent").not(':first').remove();

来源：https://stackoverflow.com/questions/30371831/jquery-remove-element-if-it-occurs-twice

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