Rotate image with javascript

Question

You use a combination of CSS's transform (with vendor prefixes as necessary) and transform-origin, like this: (also on jsFiddle)

var angle = 0,
  img = document.getElementById('container');
document.getElementById('button').onclick = function() {
  angle = (angle + 90) % 360;
  img.className = "rotate" + angle;
}

#container {
  width: 820px;
  height: 100px;
  overflow: hidden;
}
#container.rotate90,
#container.rotate270 {
  width: 100px;
  height: 820px
}
#image {
  transform-origin: top left;
  /* IE 10+, Firefox, etc. */
  -webkit-transform-origin: top left;
  /* Chrome */
  -ms-transform-origin: top left;
  /* IE 9 */
}
#container.rotate90 #image {
  transform: rotate(90deg) translateY(-100%);
  -webkit-transform: rotate(90deg) translateY(-100%);
  -ms-transform: rotate(90deg) translateY(-100%);
}
#container.rotate180 #image {
  transform: rotate(180deg) translate(-100%, -100%);
  -webkit-transform: rotate(180deg) translate(-100%, -100%);
  -ms-transform: rotate(180deg) translateX(-100%, -100%);
}
#container.rotate270 #image {
  transform: rotate(270deg) translateX(-100%);
  -webkit-transform: rotate(270deg) translateX(-100%);
  -ms-transform: rotate(270deg) translateX(-100%);
}

<button id="button">Click me!</button>
<div id="container">
  <img src="http://i.stack.imgur.com/zbLrE.png" id="image" />
</div>

var angle = 0;
$('#button').on('click', function() {
    angle += 90;
    $('#image').css('transform','rotate(' + angle + 'deg)');
});

Try this code.

No need for jQuery and lot's of CSS anymore (Note that some browsers need extra CSS)

Kind of what @Abinthaha posted, but pure JS, without the need of jQuery.

let rotateAngle = 90;

function rotate(image) {
  image.setAttribute("style", "transform: rotate(" + rotateAngle + "deg)");
  rotateAngle = rotateAngle + 90;
}

#rotater {
  transition: all 0.3s ease;
  border: 0.0625em solid black;
  border-radius: 3.75em;
}

<img id="rotater" onclick="rotate(this)" src="https://upload.wikimedia.org/wikipedia/en/e/e0/Iron_Man_bleeding_edge.jpg"/>

CSS can be applied and you will have to set transform-origin correctly to get the applied transformation in the way you want

See the fiddle:

http://jsfiddle.net/OMS_/gkrsz/

Main code:

/* assuming that the image's height is 70px */

img.rotated {
    transform: rotate(90deg);
    -webkit-transform: rotate(90deg);
    -moz-transform: rotate(90deg);
    -ms-transform: rotate(90deg);

    transform-origin: 35px 35px;
    -webkit-transform-origin: 35px 35px;
    -moz-transform-origin: 35px 35px;
    -ms-transform-origin: 35px 35px;
}

jQuery and JS:

$(img)
    .css('transform-origin-x', imgWidth / 2)
    .css('transform-origin-y', imgHeight / 2);

// By calculating the height and width of the image in the load function

// $(img).css('transform-origin', (imgWidth / 2) + ' ' + (imgHeight / 2) );

Logic:

Divide the image's height by 2. The transform-x and transform-y values should be this value

Link:

transform-origin at CSS | MDN

Hope this can help you!

<input type="button" id="left"  value="left" />
<input type="button" id="right" value="right" />
<img src="https://www.google.com/images/srpr/logo3w.png" id="image">

<script>
 var angle = 0;
    $('#left').on('click', function () {  
        angle -= 90;
        $("#image").rotate(angle);
    });

    $('#right').on('click', function () {  
        angle += 90;
        $("#image").rotate(angle);
    });
</script>

Try it

i have seen your running code .There is one line correction in your code.

Write:

$("#wrapper").rotate(angle); 

instead of:

$("#image").rotate(angle);

and you will get your desired output,hope this is what you want.

I think this will work.

    document.getElementById('#image').style.transform = "rotate(90deg)";

Hope this helps. It's work with me.