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How to position rotated element to the right edge

I am trying to align an absolute element (image in this case) to the right edge of the container.
It works if the element is not rotated, but when a transformation is involved, the left property is not calculated correctly.
Maybe I am missing something, but the solution I am using right now is getBoundingClientRect() to get the width and then subtract it from the container width.
Here is a JSFiddle that demonstrate what I am doing.

getBoundingClientRect is a good approach, the problem is that when you set css left, it positions it without the rotation calculated. The order in which you set it doesn't change the fact the the rotation is applied in relation to the css, not in relation to the current position of the rotated div. So when you calculate dimensions using getBoundingClientRect you're taking into account the rotation, then you use it on a css that doesn't take it into account.
One easy way to get proper coordinates, would be to calculate the x difference between before rotation and after and adjust you left accordingly. You'll have prevDimension.x - dimension.x giving you the difference in x that the rotation is creating, which allows you to adjust newLeft.
Like this:
$('#rotate-align').click(function () {
var prevDimensions = $('.element')[0].getBoundingClientRect();
$('.element').css('transform', 'rotate(0.99923rad)');
var dimensions = $('.element')[0].getBoundingClientRect();
var newLeft = $('#bounds').width() - dimensions.width - dimensions.x + prevDimensions.x;
$('.element').css('left', newLeft);
});
http://jsfiddle.net/jgcynwmp/3/
Another approach would be to calculate the x difference based on the width difference between the non rotated element and the rotated element. This can be done using offsetWidth (which doesn't take the rotation into account) and the getBoundingClientRect. The difference between the 2 will tell you how much width is lost with the rotation. Note that for this calculation, the transform origin is important. For example, with a centered rotation, you'll need to divide by 2 the width difference to get the x difference, but with another origin it would be something else.
Like this:
$('#rotate-align').click(function () {
$('.element').css('transform', 'rotate(0.99923rad)');
var dimensions = $('.element')[0].getBoundingClientRect();
var newLeft = $('#bounds').width() - $('.element')[0].offsetWidth + (($('.element')[0].offsetWidth - dimensions.width) / 2);
$('.element').css('left', newLeft);
});
http://jsfiddle.net/jgcynwmp/4/

There is a JSFiddle here.
When the image is rotated, the bounding rectangle remains in the place of the rotation, instead of being to the transformed coordinates.
I added a 'bcr' <div> element which then is matched to the bounding client rectangle.
After the rotation we can move the image into place (477 is the absolute right of bounds).
There appears to be a small problem if you repeatedly click the button, but I guess that's the magic of CSS transforms!
$('#align').click(function () {
var newLeft = $('#bounds').width() - $('.element').outerWidth();
$('.element').css('left', newLeft);
});
$('#rotate-align').click(function () {
$('.element').css('transform', 'rotate(0.69923rad)');
var dimensions = $('.element')[0].getBoundingClientRect();
$('.element').css('left',477-dimensions.width-dimensions.left);
$('#bcr').css('left',dimensions.left);
$('#bcr').css('top',dimensions.top);
$('#bcr').css('width',dimensions.width);
$('#bcr').css('height',dimensions.height);
});
#bounds {
width:427px;
height:354px;
left:50px;
top:38px;
border: 1px solid red;
position: absolute;
}
#bcr {
width:327px;
height:254px;
left:150px;
top:138px;
border: 1px solid green;
position: absolute;
}
.element {
top: 100px;
z-index: 102;
line-height: 82px;
width: 312px;
height: 82px;
#transform: rotate(0.99923rad);
left: 0;
position:absolute;
border: 1px solid green;
}
.element-img {
width: 100%!important;
height: 100%!important;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="bounds">
<div class="element">
<img class="element-img" src="https://www.google.com/logos/doodles/2014/2014-winter-olympics-5710368030588928-hp2x.jpg"> </div>
</div>
<input type="button" id="align" value="Align right" style="width:100%;" />
<input type="button" id="rotate-align" value="Rotate and align right" style="width:100%;" />
<div id="bcr"></div>

Margin-left animate to the end of div

I have one wrapper div (the grey background) and 5 squares inside it. After the press of a button, the blue one moves and has to stop at the end of the wrapper div, but it goes behind it. How do I make it go to the end of the div, and not behind it?
There is what I've tried so far:
<button id = "start">
Start
</button>
<div style="background-color:rgb(201, 201, 201);width:80%;height:250px" id="horsewrapper">
<div style="height: 50px; width: 100px; text-align: center; background-color: blue;" id="horse1">1</div>
<div style="background-color:red;text-align:center;height:50px;width:100px" id="horse2">1</div>
<div style="background-color:green;text-align:center;height:50px;width:100px" id="horse3">1</div>
<div style="background-color:yellow;text-align:center;height:50px;width:100px" id="horse4">1</div>
<div style="background-color:orange;text-align:center;height:50px;width:100px" id="horse5">1</div>
</div>
Demo can be found here:
https://jsfiddle.net/wqrun6ny/2/
Thanks

https://jsfiddle.net/wqrun6ny/3/
Margin 100% adds margin to widh of element. To avoid this you should add to your animate function left property which is equal to width of element:
$('#horse1').animate({"margin-left":"100%", 'left': -100} ....
but it will works only if element has position:relative

You can make a calculation before start the animation. It takes the width of the wrapper and substract it the width of the "horse":
https://jsfiddle.net/wqrun6ny/4/
$('#start').click(function(){
var margin = $('#horsewrapper').width() - $('#horse1').width();
$('#horse1').animate({"margin-left": margin},{"duration":1000,"easing":"linear"});
});
Edit
According with the request in comments, you can use stop()method and then reinitialise the animation, it works perfectly:
https://jsfiddle.net/wqrun6ny/15/
$('#start').click(function(){
animate($('#horse1'));
});
$(window).on('resize', function() {
$('#horse1').stop();
animate($('#horse1'));
});
var animate = function(element) {
var margin = $('#horsewrapper').width() - element.width();
element.animate({"margin-left": margin},{"duration":5000,"easing":"linear"});
};
You will notice a problem, if you don't push the button but you resize the window, it will start the animation. To avoid this you can add a flag or check if the div is in the initial position.

You have to calculate margin first, then animate according to margin.
Try like this:
$('#start').click(function(){
var mar = $('#horsewrapper').width() - $('#horse1').width();
$('#horse1').animate({"margin-left": mar}, {"duration":1000,"easing":"linear"});
});

https://jsfiddle.net/wqrun6ny/14/
You could calculate the width of the container, and subtract the width of the boxes your moving. Pushing it 100% will result in what your demo is displaying.
If its a static box width the same width, you can just use a static pixel value as well.
var horsewrapperWidth = $('#horsewrapper').width() -100;
100 is the width of your "horses".
$('#horse1').animate({"margin-left": horsewrapperWidth + 'px'},{"duration":5000,"easing":"linear"});

Simplest way !
$('#start').click(function() {
// added this variable to get width of box
var box = $('#horsewrapper').width();
$('#horse1').finish().css("margin-left", "initial");
$('#horse1').animate({
"margin-left": box - 100 // box - width of horse (100%-100px)
}, {
"duration": 5000,
"easing": "linear"
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button id="start">
Start
</button>
<div style="background-color:rgb(201, 201, 201);width:80%;height:250px" id="horsewrapper">
<div style="height: 50px; width: 100px; text-align: center; background-color: blue;" id="horse1">1</div>
<div style="background-color:red;text-align:center;height:50px;width:100px" id="horse2">1</div>
<div style="background-color:green;text-align:center;height:50px;width:100px" id="horse3">1</div>
<div style="background-color:yellow;text-align:center;height:50px;width:100px" id="horse4">1</div>
<div style="background-color:orange;text-align:center;height:50px;width:100px" id="horse5">1</div>
</div>

How to determine element coordinates in scrollable div?

I have a div element with vertical scrolling. It has span elements with text in it. How can I get the coordinates of these span elements. But I don't want it relative to the scroll position.
For example, in the div, lets say its 400px width, 1000px in height (but the view height is 500px), and its scrolled half way vertically. Then in the center of the view, I see a text, and if I click on it, I want the coordinate like (200, 250)
How can I get coordinates that are absolute to the div container?

Use element.getBoundingClientRect? (Subtract the container's top if you want it relative to the container.)
function client() {
alert(document.getElementById('child').getBoundingClientRect().top);
}
function local() {
var container = document.getElementById('container');
var containerRect = container.getBoundingClientRect();
var child = document.getElementById('child');
var childRect = child.getBoundingClientRect();
localTop = childRect.top - containerRect.top;
alert(localTop);
}
#container {
height: 100px;
overflow:scroll;
border:1px solid red;
margin-top:100px;
}
#child {
background: yellow;
}
<div id="container">
<div>Test</div>
<div>Test</div><div>Test</div><div>Test</div><div>Test</div>
<div id="child">Item of interest</div>
<div>Test</div><div>Test</div><div>Test</div><div>Test</div>
</div>
<button onclick="client()">Global top</button>
<button onclick="local()">Local top</button>

Let's say the scrollable div has an id of 'myDiv' and one element inside the div has an id of 'element1'. To find its true 'top' value in javascript try this:
alert($('myDiv').css('top') + $('element1').css('top'));

How to find a position somewhere in div tag using jquery

Also please view the attached image for clarification. I have a div container what I want to to find a position somewhere in that div container using jquery or javascript or both. The attached image shows everything. Please help.
Update
The reason I want to find this position is that I want to animate container towards that point and eventually disappear. Secondly I would like to find position on the opposite side too so that I could animate container from that position.
Second update
In other words how can we find the point of intersection of two lines?

Given you need to find the intersection between two lines inside a div, your markup could look like this:
<div id="container" style="position:absolute; width: 100%; height: 200px;">
<div style="width: 2px; height: 100%; left: 20%; position:absolute; background-color: red; top: 0;"></div>
<div style="height: 2px; width: 100%; left: 0; position:absolute; background-color: blue; top: 25%;"></div>
</div>​
Using jQuery, you can find the coordinates for the intersection like this:
var x = $('#container div:first').position().left;
var y = $('#container div:last').position().top;
console.log(x,y);
x and y would be the coordinates in pixels relative to the container element.
http://jsfiddle.net/sAsmj/

I dont see the image, however if you are looking at getting the position, which is ideally caret position, you can use the jquery plugin http://plugins.jquery.com/project/jCaret

You can find the poiter position by using this, try it
$(document).ready(function(){
$("div#container").on("mousemove", function(e){
var self = $(this);
var dx = e.pageX;
var dy = e.pageY;
var x = dx - self.offset().left ;
var y = dy - self.offset().top ;
console.log(x);
console.log(y);
});
});

If you want the X, Y of the mouse you can read this question:
getting the X/Y coordinates of a mouse click on an image with jQuery
Here is an excerpt from the question which is based upon an img but you can change it for your container:
$(document).ready(function() {
$('img').click(function(e) {
var offset = $(this).offset();
alert(e.clientX - offset.left);
alert(e.clientY - offset.top);
});
});

Resize based on page/screen height

My page is divided into left and right divs, the right div has a border left partitioning the two. if the height of the right box is bigger then left, it works fine. However if the left box height is more, then the border is only halfway.
How can i resize the height of the right box based on the height of entire screen so that the border runs all the way to the end.

You can provide height to your right div like, place a id ( like rightDiv ) there if not (in jQuery).
$('#rightDiv').height($(window).height());
if you want to height of your entire document use:
$('#rightDiv').height($(document).height());
$(window).height() will retrun available browser window height.
$(document).height() will retrun document height.
or you can make a comparison:
var doc = $(document);
var win = $(window);
var maxHeight = doc.height() > win.height() ? doc.height() : win.height() ;
$('#rightDiv').height(maxHeight);
You have min-height, for animate height you can try:
$('#rightDiv').animate( { height : maxHeight}, <duration>);
<duration> is optional, you can provide here 'slow', 'fast', miliseconds

Another solution would be this pure CSS one: http://jsfiddle.net/zgMv5/
You put around the left and the right div another <div> and use it as CSS table row. Then the 2 containing <div> will be the same height.
<div id="outer">
<div id="left">This is some text.</div>
<div id="right">This is some text.</div>
</div>
The corresponding CSS would look like this:
div#outer {
display:table-row; }
div#outer > div {
display:table-cell; }
div#left {
border-right:1px solid red; }
I am not sure about the compatibility with old browsers...