JS / jQuery：隐藏没有特殊属性值的元素

Question

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

The code below will effectively hide all elements where attribute picturename contains "horse":

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

But how do I hide all elements where attribute picturename DOESNT contain "horse"? You would think adding a ! before the = would do the trick but no.

Answer 1

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

这应该做。

Answer 2

Alternative you can use:

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

 <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> 

After your last comment to hide when doesn't contain horse or dog you can use:

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