CSS Square Div with an Inward Oval Shape - html

Question

I am trying to create a div in css with an inward oval shape to it like this.
At the moment, I have a shape that is outward instead of inward (JS Fiddle Link).
.shape {
float: left;
width: 100px;
height: 50px;
border: none;
background: #CC0000;
border-radius: 0 90px 0 0;
-moz-border-radius: 0 90px 0 0;
-webkit-border-radius: 0 90px 0 0;
background-image: -webkit-gradient( linear, left top, right bottom, color-stop(0, #520C0C), color-stop(1, #CC0000) );
background-image: -o-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -moz-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -webkit-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -ms-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: linear-gradient(to right bottom, #520C0C 0%, #CC0000 100%);
}
Any ideas on how to go about this?

Answer 1

Have a look at my example fiddle.
I used a pseudo-element and some elliptical border-radius coupled with an inset box-shadow.
div {
position:relative;
width: 200px;height: 100px;
background: #CC0000;
background-image: linear-gradient(to right bottom, #520C0C 0%, #CC0000 100%);
}
div:after {
position:absolute;content:"";
width: 100%;height: 95%;
background: #222;
box-shadow:inset 10px -10px 5px -10px #000;
border-radius: 0 0 0 200px / 100px;
}
With a little more effort, one could probably get closer to your result, but this might be a good starting point.

Answer 2

I have created this fiddle for you. Here is the code:
HTML
<div class="container">
<div class="shape"></div>
</div>
CSS
.container {
float: left;
width: 100px;
height: 50px;
border: none;
background: #CC0000;
background-image: -webkit-gradient( linear, left top, right bottom, color-stop(0, #520C0C), color-stop(1, #CC0000) );
background-image: -o-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -moz-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -webkit-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -ms-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: linear-gradient(to right bottom, #520C0C 0%, #CC0000 100%);
}
.shape {
width: 100px;
height: 50px;
border: none;
background: #000000;
border-radius: 0 0 0 90px;
-moz-border-radius: 0 0 0 90px;
-webkit-border-radius: 0 0 0 90px;
}

Answer 3

If the part of the graphic that "isn't there" doesn't have to be actually transparent, then you can just make a regular rectangle, and build a curved shape that will sit on top of the rectangle and has the same color as the background.
http://jsfiddle.net/ub8fM/1/
.shape {
float: left;
width: 100px;
height: 50px;
border: none;
background: #CC0000;
background-image: -webkit-gradient(linear, left top, right bottom, color-stop(0, #520C0C), color-stop(1, #CC0000));
background-image: -o-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -moz-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -webkit-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: -ms-linear-gradient(right bottom, #520C0C 0%, #CC0000 100%);
background-image: linear-gradient(to right bottom, #520C0C 0%, #CC0000 100%);
position:relative;
}
.shape:before {
border-radius: 0 90px 0 0;
-moz-border-radius: 0 90px 0 0;
-webkit-border-radius: 0 90px 0 0;
content:'';
position:absolute;
top:0;
left:0;
background:white;
width:100%;
height:100%;
}
Having the shadow would a bit harder and I have no solution for that yet.
Also jsfiddle has a tidy up button that's super useful.