Find div which does not contain a specific property under specific attribute using CSS selector

Question

I have the following situation:

<div id="foo">
   <div id="1" style="transform:translate3d(5px,5px,5px);"></div>
   <div id="2" style="background-color:black;"></div>
</div>

I want to find the div that does not have the 'transform' property ie div with 'id = 2'.

I have tried the following, but it only selects the div having transform property, I want the exact opposite.

$('#foo').find('div[style*="transform"]');

Answer 1

 $('#foo').find('div:not([style*="transform"])').css('color','red') 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="foo"> <div id="1" style="transform:translate3d(5px,5px,5px);">1</div> <div id="2" style="background-color:black;">2</div> </div> 

1. Use :not()

Answer 2

这很好

$('#foo div').not('div[style*="transform"]');

Answer 3

 console.log($('#foo').find('div:not([style*="transform"])').text()); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="foo"> <div id="1" style="transform:translate3d(5px,5px,5px);">111</div> <div id="2" style="background-color:black;">222</div> </div>