Getting started with Twitter Bootstrap, I ran into a strange problem of which I can’t figure out the cause.
When I render an image using:
<img class="img-rounded" id="logo" />
and in CSS:
#logo {
background-image: url(/Content/Images/logo.png);
}
The image is shown with a narrow ‘border’:
Even though I can’t find anything remotely related to this effect. The corners are open because of the img-rounded
class.
Rendering the image using:
<img class="img-rounded" id="logo" src="~/Content/Images/SO.png" />
Renders as expected:
How can I get rid of the border?
CSS code I’ve tried without success:
border: none;
color: blue;
background-color: blue;
border-style: none;
5
Answers
The answer is not to use an
<img>
tag and try to set abackground-image
in css.Either use a
<div>
or<img src="pic.png">
and not an invalid mix of both.Credits to rblarsen who pointed this out.
using the image as a background. Why not set it as the src property of the button
When you set the type as image the input expects an src attribute as well..
Reference: http://www.w3.org/TR/html401/interact/forms.html#adef-src and http://www.w3.org/TR/html401/interact/forms.html#h-17.4.1
Just add this, hope it’ll work well
This should work. I believe the problem is to do with the
img
tag used without thesrc
attribute. In that case, we can simply use adiv
tag.I have come across this problem earlier on SO, I do not remember the link to that question though.
1) Make your
height
andwidth
as 0.2) Give appropriate padding as per the image size. (i.e padding: width/2px height/2px width/2px height/2px )
In short your left and right padding should add upto
width
of the imageAND
your top and bottom padding should add upto
height
of the image.For people running into this situation, the actual solution to the problem in question (as opposed to people telling you “just don’t do that”) is to set the
src
of the image to a blank image; preferably a1x1
image to reduce load size, and optionally even as an incorporatedbase64
URI to eliminate the additional HTTP request (depending on which solution is more appropriate).So from the OP’s example, the altered code:
or