如何獲取所選元素后的所有元素

Question

如何包含最后一個元素？

 $('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

您可以使用nextAll()代替：

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

 .foo { color: #00F; } 

 <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 2

您可以將andSelf方法與next()方法結合使用。

andSelf方法將堆棧上的前一組元素添加到當前集合。

 $('div.current').nextUntil("div:last").andSelf().next().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 3

您可以在選擇器中使用通用兄弟組合器~

 $('div.current ~ div').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 4

你可以在nextUntil()傳遞沒有參數，它將選擇所有下一個元素。

或者也許使用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。

 $('div.current').nextAll().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 6

這是你想用你的代碼做的^^

 $('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>