如何使用jQuery选择具有属性和属性不等于特定值的元素？

Question

I got the following line:

$("element").find("[attr!='val'][attr!='val'][attr!='val'][attr!='val'][attr!='val']") 

based on this answer. My page stops running. My assumption is that it finds all the different elements in the page, also the ones that don't have the attribute. Is that correct? If so how can I fix this? I tried the

$("element").not("[attr='val']") 
// instead of 
$("element").find("[attr!='val']")

and it still crashed the page. The following code works of course:

$("element").find("[attr='val']")

Answer 1

JQuery Has Attribute Selector [name] select element only has attribute without consider to it value. You need to adding [attr] at the first of your selector.

 $("ul").find("[class][class!='A']").css("color", "red"); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <ul> <li class="A">A</li> <li class="A">A</li> <li class="B">B</li> <li>C</li> </ul>