JQuery：如何删除除第一个包含子字符串为'id'标记之外的所有div元素？

Question

I have few div elements having id as 'name_a_1', 'name_a_2', 'name_b_3' so on. I want to remove all divs except the first name, which starts from 'name_a_' using JQuery or plain JS?

I successfully tried accessing first and last elements by following JQuery, but how can I access all elements except first?

$('[id^="name_a_"]').first()
$('[id^="name_a_"]').last()

Answer 1

You can use combination :not() and :first selector to target all element except first.

$('[id^="name_a_"]:not(:first)').remove()

or, :gt(index) selector

Select all elements at an index greater than index within the matched set.

$('[id^="name_a_"]:gt(0)').remove()

---

To check if element have value, .filter() can be used

var hasValue = $('[id^="name_a_"]').filter(function(){
    return $(this).val().trim() !== '';
}).length > 0;

Answer 2

Plain JS example using Array.prototype.slice to get all matching elements except for the first.

 Array.prototype.slice.call(document.querySelectorAll('[id^="name_a_"]'), 1) .forEach(elt => elt.parentNode.removeChild(elt)) 

 <ul> <li id="name_a_1">name_a_1</li> <li id="name_a_2">name_a_2</li> <li id="name_a_3">name_a_3</li> <li id="name_a_4">name_a_4</li> </ul> 

Answer 3

您还可以使用.not排除第一个div：

$('div[id^=name_a]').not(':eq(0)').remove();