Question

如何包含最后一个元素？

$('div.current').nextUntil("div:last").css("color", "blue");

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<div>First</div>
<div>Second</div>
<div class="current">Third</div>
<div>Fourth</div>
<div>Fifth</div>
<div>Sixth</div>
<div>Seventh</div>

Answer 1

您可以改为使用string1 <TAB> string2 <TAB> 9999 abc dehi [way:pn9999] <TAB> 001 <TAB> org; string3 string4 string5 <TAB> string6 <TAB> 9999 dwd meti [way:pn8999] <TAB> 002 <TAB> org2; string7 string8 <TAB> string9 <TAB> 9999 dwd meti [way:pn7999] <TAB> 003 <TAB> org4; string10：

＆＃13;

nextAll()

＆＃13;

$('div.current').nextAll().addClass('foo');

＆＃13;

.foo { color: #00F; }

＆＃13;

Answer 2

您可以将andSelf方法与next()方法结合使用。

andSelf方法将堆栈上的前一组元素添加到目前的设定。

＆＃13;

$('div.current').nextUntil("div:last").andSelf().next().css("color", "blue");

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<div>First</div>
<div>Second</div>
<div class="current">Third</div>
<div>Fourth</div>
<div>Fifth</div>
<div>Sixth</div>
<div>Seventh</div>

＆＃13;

Answer 3

您可以在选择器中使用general sibling combinator ~

＆＃13;

$('div.current ~ div').css("color", "blue");

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<div>First</div>
<div>Second</div>
<div class="current">Third</div>
<div>Fourth</div>
<div>Fifth</div>
<div>Sixth</div>
<div>Seventh</div>

＆＃13;

Answer 4

您可以在nextUntil()内部不传递参数，它将选择以下所有元素。

或者也许使用nextAll()。

$('div.current').nextUntil().css("color", "blue");

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<div>First</div>
<div>Second</div>
<div class="current">Third</div>
<div>Fourth</div>
<div>Fifth</div>
<div>Sixth</div>
<div>Seventh</div>

Answer 5

使用jQuery的nextAll方法而不是nextUntil。

＆＃13;

$('div.current').nextAll().css("color", "blue");

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<div>First</div>
<div>Second</div>
<div class="current">Third</div>
<div>Fourth</div>
<div>Fifth</div>
<div>Sixth</div>
<div>Seventh</div>

＆＃13;

Answer 6

以下是您要对代码执行的操作^^

$('div.current').nextUntil("html").css("color", "blue");

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<div>First</div>
<div>Second</div>
<div class="current">Third</div>
<div>Fourth</div>
<div>Fifth</div>
<div>Sixth</div>
<div>Seventh</div>

如何获取所选元素后的所有元素

6 个答案: