linear-gradient(…) in CSS3 not working according to specifications?
Question
Here's the current version of my webpage:
<html>
<head>
<title>my site</title>
<style type="text/css">
#mtx_bckgd
{
font-family: Courier New;
height: 1000px;
width: 1000px;
position: absolute;
z-index: -1;
}
#mtx_bckgd > p
{
word-wrap: break-word;
color: #D8D8D8;
overflow: hidden;
}
#header
{
position: absolute;
font-family: Trebuchet MS;
width: 1000px;
height: 70px;
text-align: center;
border: solid 2px #424242;
top: 25px;
font-size: 20;
}
</style>
<script type="text/javascript">
function change_bckgd()
{
var bitstr = "";
for (var i = 0; i < 4000; ++i)
bitstr += Math.floor(Math.random()*10) % 2 ? "0" : "1";
document.getElementById("mtx_txt").innerHTML = bitstr;
}
</script>
</head>
<body>
<div id="mtx_bckgd">
<p id="mtx_txt"></p>
</div>
<div id="header">
<h1>JaminWEB</h1>
</div>
<script type="text/javascript">
setInterval(change_bckgd, 200);
</script>
</body>
</html>
What I plan on doing is creating a vertical fade across the background, from white at the bottom to light grey (the same grey I used for the 0's and 1's in the background) at the top.
According to W3schools ( http://www.w3schools.com/css/css3_gradients.asp ), the syntax
background: linear-gradient(angle, color-stop1, color-stop2);
means I should add the line
background: linear-gradient(90, white, #D8D8D8)
in the #mtx_bckgd block in the stylesheet. But when I do that, I'm getting a dark blue fade and my 0's and 1's are turning white.
1 answers
solution1
1 ACCPTED 2014-03-20 21:55:25
The problem is the "90" in background: linear-gradient(90, white, #D8D8D8) is incorrect. According to W3Schools it should be:
background: linear-gradient(90deg, white, #D8D8D8)
or
background: linear-gradient(to right, white, #D8D8D8);
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