CSS Border rotation [duplicate]

Question

This question already has an answer here:

• Shape with a slanted side (responsive) 2 answers

In css can i rotate the border alone, instead of rotating the whole element? something like this:

I basically wanna create a slanting border for my video player.

I wanna do something like the accepted answer of this post: click here

except that instead of slanting the top and bottom it slants only the right side.

I've tried this, but it slants both left and right sides:

html:

<div class="skew-neg">
    <p>Hello World.</p>
    <p>My name is Jonathan.</p>
    <p>This box is skewed.</p>
    <p>In supported browsers.</p>
</div>​

css:

html { 
    background: #FFF;
    color: lightblue;
    font: 16px 'Arial';
    line-height: 150%;
}

div {
    background: blue;
    margin: 50px auto 0;
    padding: 20px;
    width: 200px;
    box-sizing: border-box;
    box-shadow: 0 0 20px rgba(0,0,0,.9);
    border-radius: 25px;
}

.skew-neg {
    -webkit-transform: skewX(-50deg);
    -moz-transform: skewX(-50deg);
    -ms-transform: skewX(-50deg);
    -o-transform: skewX(-50deg);
    transform: skewX(-50deg);
}

.skew-neg > * {
    -webkit-transform: skewX(50deg);
    -moz-transform: skewX(50deg);
    -ms-transform: skewX(50deg);
    -o-transform: skewX(50deg);
    transform: skewX(50deg);
}

Answer 1

Follow here to solve this problem :)

click here

Hopefully it can help you

.b-border{
    display: inline-block;
    position: relative;
    border: solid #333;
    border-width: 1px 1px 0px 1px;
    padding: 20px;
    margin-bottom: 100px;
}
.b-border.border-right{
  border-width: 1px 0 1px 1px;
}

.b-border.border-right:after{
    content: "";
    position: absolute;
    right: -30px;
    border-top: 1px solid #333;
     border-left: none medium;
    top: -1px;
    left: auto;
    width: 30px;
    height: 100%;
    background: linear-gradient(to top left, white calc(50% - 1px), #000 1px, white 50%);
}
.b-border:after{
    content: "";
    position: absolute;
    left: -1px;
    bottom: -15%;
    border-left: 1px solid #333;
    height: 15%;
    width: calc(100% + 1px);
    background: linear-gradient(to right bottom, white calc(50% - 1px), #000 1px, white 50%);
}

<div class="b-border border-right">
    Hello :)
</div>
<p></p>
<div class="b-border" style="width: 300px;">
    Lorem ipsum dolor sit amet, consectetur 
    Lorem ipsum dolor sit amet, consectetur
    Lorem ipsum dolor sit amet, consectetur
     Lorem ipsum dolor sit amet, consectetur 
     Lorem ipsum dolor sit amet, consectetur 
</div>

Answer 2

What you can do is something like this: http://jsfiddle.net/py9Ze/

HTML:

<div id="vid-container">
    <span><embed>this is straight</embed></span>
</div>

CSS:

#vid-container {
    border: 1px solid;
    height: 200px;
    -moz-transform: scale(1) rotate(10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    -webkit-transform: scale(1) rotate(10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    -o-transform: scale(1) rotate(10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    -ms-transform: scale(1) rotate(10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    transform: scale(1) rotate(10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
}
span {
    position: absolute;
    -moz-transform: scale(1) rotate(-10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    -webkit-transform: scale(1) rotate(-10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    -o-transform: scale(1) rotate(-10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    -ms-transform: scale(1) rotate(-10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
    transform: scale(1) rotate(-10deg) translateX(0px) translateY(0px) skewX(0deg) skewY(0deg);
}

Basically put a border around the parent container and rotate it x degrees, then rotate the child -x degrees.

PS I am behind a firewall so I cant see the image you posted so this might be way off from what you wanted.

Answer 3

You could try using CSS Generated Arrows. Here is an overview of how to create and use them.

I don't believe this is the solution, BUT it could be one of them. You might find the trick helpfull.

Best

Answer 4

A solution that require JavaScript and canvas, but offers great versatility -

Result:

ONLINE DEMO

Code:

function makeBorder(id, bw, rSkew, radius) {

    var el = document.getElementById(id),
        canvas = document.createElement('canvas'),
        ctx = canvas.getContext('2d'),

        bwh = bw / 2,
        w = parseInt(getComputedStyle(el).getPropertyValue('width'), 10),
        h = parseInt(getComputedStyle(el).getPropertyValue('height'), 10);

    canvas.width = w;
    canvas.height = h;

    /// draw border        
    ctx.beginPath();
    roundedRect(ctx, bwh, bwh, w - bwh, h - bwh, radius, rSkew);
    ctx.lineWidth = bw;
    ctx.stroke();

    /// set as background
    el.style.background = 'url(' + canvas.toDataURL() + ') no-repeat top left';
}

The add this for creating the rounded rectangle (with mod. for skew):

function roundedRect(ctx, x, y, w, h, rul, skew) {
    //modification to fit purpose here

    var rur = rul,
        rbr = rul,
        rbl = rul,
        dul = rul * 2,
        dur = rul * 2,
        dbr = rul * 2,
        dbl = rul * 2,
        _x, _y,
        ww = x + w,
        hh = y + h,
        rr,
        pi = Math.PI,
        pi15 = Math.PI * 1.5,
        pi05 = Math.PI * 0.5;

    //Upper Left    
    rr = [x, y, dul, dul];
    _x = rr[0] + rr[2] / 2;
    _y = rr[1] + rr[3] / 2;
    ctx.arc(_x, _y, rul, pi, pi15);

    //Upper right
    rr = [ww - dur, y, dur, dur];
    _x = rr[0] + rr[2] / 2;
    _y = rr[1] + rr[3] / 2;
    ctx.arc(_x, _y, rur, pi15, 0);

    ctx.lineTo(ww - skew, h);

    //Bottom left
    rr = [x, hh - dbl, dbl, dbl];
    _x = rr[0] + rr[2] / 2;
    _y = rr[1] + rr[3] / 2;
    ctx.arc(_x, _y - 1, rbl, pi05, pi);
    ctx.closePath();
}

Then you just call this function with ID of element, border width and how many pixels you want to skew the right side with:

makeBorder('demo', 2, 50, 7);