Creating a DIV that uses CSS to draw a triangle to the left. Trying to apply a uniform box-shadow to both parent and the pseudo element (see images) and code.
Is this possible? Or am I better off using border-image for this?
(Top: Before Shadow, Middle: CSS Box-Shadow, Bottom: Desired Result)
.bubble{
height: 200px;
width: 275px;
opacity: 0;
margin-top: 41px;
float: right;
background-color: #F2F2F2;
-webkit-border-radius: 5px;
-webkit-box-shadow: 0px 0px 6px #B2B2B2;
}
.bubble::after {
height: 0px;
width: 0px;
content: "\00a0";
display: block;
margin-left: -10px;
margin-top: 28px;
border-width: 10px 10px 10px 0;
border-style: solid;
border-color: transparent #F2F2F2 transparent transparent;
-webkit-box-shadow: 0px 0px 6px #B2B2B2;
}
Instead of using a triangle hack, you can just rotate a div using transform
and get a real box-shadow
. Since you only want the shadow on one side of the div (the visible triangle side), you have to make the blur
smaller and lower the opacity
.
Demo: http://jsfiddle.net/ThinkingStiff/mek5Z/
HTML:
<div class="bubble"></div>
CSS:
.bubble{
background-color: #F2F2F2;
border-radius: 5px;
box-shadow: 0px 0px 6px #B2B2B2;
height: 200px;
margin: 20px;
width: 275px;
}
.bubble::after {
background-color: #F2F2F2;
box-shadow: -2px 2px 2px 0 rgba( 178, 178, 178, .4 );
content: "\00a0";
display: block;
height: 20px;
left: -10px;
position: relative;
top: 20px;
transform: rotate( 45deg );
-moz-transform: rotate( 45deg );
-ms-transform: rotate( 45deg );
-o-transform: rotate( 45deg );
-webkit-transform: rotate( 45deg );
width: 20px;
}
Output: