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4048

answers:

8

I know that Internet Explorer has some proprietary extensions so that you can do things like create divs with a gradient background. I can't remember the element name or it's usage. Does anyone have some examples or links?

+3  A: 

Look at the custom CSS filters IE can handle http://msdn.microsoft.com/en-us/library/ms532847.aspx

Nick
That's it! thanks.
Jeremy
A: 

Is this what you are looking for? IE Gradients

Jeremy B.
Dead link!
Dimitri C.
A: 

I believe you're looking for this specific CSS setting.

tsilb
A: 

filter: progid:DXImageTransform.Microsoft.Gradient( gradientType=0,startColorStr=#d8cde9, endColorStr=#8f7eac);

Liam
A: 

Right from ScriptFX.com article:

<body bgcolor="#000000" topmargin="0" leftmargin="0">

    <div style="width:100%;height:100%; filter: progid:
        DXImageTransform.Microsoft.Gradient (GradientType=1,
        StartColorStr='#FF006600', EndColorStr='#ff456789')">

Your page content goes in here ...... at the end of all the page content, you must close the <div> tag, immediately before the closing <body> tag.... as below

    </div>
</body>
vmarquez
+1  A: 

This should work for IE6, IE7, and IE8:

.gradientClass
{
  filter:  progid:DXImageTransform.Microsoft.gradient(GradientType=0,startColorstr='#e6e6e6', endColorstr='#CCCCCC'); /* IE6 & IE7 */
  -ms-filter: "progid:DXImageTransform.Microsoft.gradient(GradientType=0,startColorstr='#e6e6e6', endColorstr='#CCCCCC')"; /* IE8 */
}
James Lawruk
A: 

The code I use for all browser gradients..

        background: #0A284B;
        background: -webkit-gradient(linear, left top, left bottom, from(#0A284B), to(#135887));
        background: -moz-linear-gradient(top, #0A284B, #135887);
        filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#0A284B', endColorstr='#135887');

Gradients don't work on ie <li></li> elements

Blowsie
You will ned to specify a height for the element for this to work in ie
Blowsie
A: 

hi Blowsie, I am using IE8 and the gradient you specified for IE is not working in my case.Can you help me in solving this problem.

thanks in advance

ganesh