JS / jQuery：隐藏没有特定属性值的元素

Question

<div class="picture" picturename="picture of a horse and stuff"></div>

以下代码将有效隐藏属性picturename包含的所有元素＆＃34; horse＆＃34;：

$('[picturename*="horse"]').hide();

但是如何隐藏属性picturename DOESNT包含的所有元素＆＃34; horse＆＃34;？ 你会想添加一个！在=之前会做的但是没有。

Answer 1

$('.picture').not('[picturename*="horse"]').hide();

应该这样做。

Answer 2

您可以使用替代方案：

＆＃13;

＆＃13;

$('.picture:not([picturename*="horse"])').hide();

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="picture" picturename="picture of a horse and stuff">show</div>
<div class="picture" picturename="picture of a horse and stuff">show</div>
<div class="picture" picturename="picture of a and stuff">hide</div>
<div class="picture" picturename="picture of a horse and stuff">show</div>

＆＃13;

＆＃13;

＆＃13;

在你最后的评论中隐藏，当你不包含马或狗时，你可以使用：

$('.picture:not([picturename*="horse"]):not([picturename*="dog"])').hide();