How can I go about positioning an element in the same relative position in the viewable area of the window regardless of the window size?
For example, an element that should be:
30% of the window width from the left of the page, and
50% of the window height from the top of the page.
In most cases it's possible to do this with CSS only. Did you try to use position: fixed?
You can use absolute positioning and viewport percentage lengths, where one vh unit is 1% of the height of the viewable area and one vw unit is 1% of the width of the viewable area.
Viewport-percentage lengths define the value relative to the size of the viewport, i.e., the visible portion of the document.
- https://developer.mozilla.org/en-US/docs/Web/CSS/length#Viewport-percentage_lengths
.demo {
position: absolute;
top: 50vh;
left: 30vw;
background: #AFAFAF;
}
<div class="demo">Hello World!</div>
Related
So I want to have a central div that is full screen, and to make the width and height of the body 150 vw and vh, with a margin of 50% so when you load the page it holds the div central, but you can scroll up, down, left, right, outside of this central div a little bit.
If you do the div 100vw, 100vh, and then the body 200vw, 200vh, it only enables scrolling to the right and downwards.
This is my understanding of how to get towards what I'm trying to do:
HTML:
<div>
centered and full screen div?
</div>
</body>
CSS:
body{
width:150vw;
margin-left:50vw;
height:150vh;
margin-top:50vh;
}
div{
width: 100vw;
height: 100vh;
border: solid;
}
JAVA:
window.scrollTo(50 + 'vw', 50 + 'vh');
https://jsfiddle.net/dsLnzyxw/3/
But this doesnt work as the javascript doesn't accept vw in the scrollTo function.. but just to give a better idea of what I'm trying to do.
Also understand I might be going around on crazy route trying to achieve something that could be done quite simple in css?
How do I achieve this?
Thanks !
In order to get units relative to the viewport width and viewport height (vw and vh) you could do simple calculations:
window.scrollTo(0.5 * window.innerWidth, 0.5*window.innerHeight);
which would set the scroll position to 50% of the window width and 50% of the window height. window.innerWidth returns the width of the window, and this is multiplied by 50% to get 50% of the width of the window in pixels.
The same goes for height. Setting the scroll position is not possible without JavaScript unfortunately.
I have a responsive background image with a smaller image positioned over it. I am trying to keep the smaller image at a specific location when the window is resized.
Both images scale properly, and the left position works so far, but not the top position.
img {
max-width:100%;
}
#dot {
position: absolute;
top: 17%;
left: 66.5%;
width: 10%;
height: 0;
padding-bottom: 10%;
}
I have found some questions with answers that suggest:
Vertical Alignment or Positioning with Javascript
I've also looked into .position() and .offset(), not sure if either would work.
I think my best solution would be to calculate the Y offset using the current window height as a reference but I am not sure what my JS or Jquery code should look like.
Here is my jsfiddle:
http://jsfiddle.net/melissadpelletier/xBu79/21/
I'm not sure exactly what you're trying to do with your images, but you could create a new smaller image (green dot) with the same aspect ratio as your background image, and have the dot placed where it needs to be within that aspect ratio. Then stretch the width of that to be 100% and the two images are basically overlapping, but the top image (smaller image) has a transparent background. Not sure if that all makes sense, but I made a new image and did the fiddle thing, which I'm new to: http://jsfiddle.net/ydack/
img
{
width:100%;
}
#dot
{
width: 100%;
position: absolute;
top:0;
left:0;
}
#dotImg
{
top: 0;
left: 0;
width: 100%;
}
I mistakenly placed the green dot's position based on the black outline, not the full background image, so the dot is slightly up and right of where it needs to be. BUT, the position is maintained while re-sizing the window. Hacky, but it could work!
You are definitely gonna need some javascript for this. What you can do is calculate the height and width of the image whenever you resize your browser window. Then simply use some math to calculate the position of the dot relative to those dimensions.
var height = $('#image').height();
var width = $('#image').width();
/* change the fractions here according to your desired percentages */
$('#dot').css({left: width/2, top: height/2});
$(window).resize(function() {
height = $('#image').height();
width = $('#image').width();
/* change the fractions here according to your desired percentages */
$('#dot').css({left: width/2, top: height/2});
});
Try this code: http://jsfiddle.net/LimitedWard/FFQt2/3/
Note that you will need to also resize the dot according to the height/width of the image if you want it to always fit inside that box.
Edit: after further investigation, it is possible to do this in CSS; however, it's a lot sloppier because the dot doesn't follow the image if the window is too wide. This jQuery solves that problem by using pixel-based positioning.
http://jsfiddle.net/sajrashid/xBu79/24/
plenty of errors mainly not closing tags
<div id='background'>
<img src='http://i.imgur.com/57fZEOt.png'/>
<div id='dot'>
<img src='http://i.imgur.com/yhngPvm.png'/>
</div>
</div>
I am trying to create a responsive design for my app.
I have a big background image and it will show the full size when user has large screen and only show partial if user uses small screen.
I want to place few elements on my app with absolute position but the problem is I can't lock their top and left value because the screen size changes.
I have something like
css
#background{
background: url('BG.jpg') no-repeat top center fixed;
width: 1900px;
height: 1200px;
}
#element{
position: fixed;
z-index: 5;
top: 50%; //looks fine in 1900 x 1200 screen but position is off on 1200 x 1000
left:70%; //looks fine in 1900 x 1200 screen but position is off on 1200 x 1000
}
html
<div id='background'></div>
<img id='element' src='test.jpg' />
How do I keep the position of the element on the same spot when user has small screen? Thanks for the help!
When using position: absolute, you need to make sure that it has a parent with a position attribute other than the default (which is static). If there is no such parent, the document is the effective parent. For your example, I would advise making the img#element a child of div#background like so
<div id='background'>
<img id='element' src='test.jpg' />
</div>
and then adding position:relative; to the #background css style
#background{
background: url('BG.jpg') no-repeat top center fixed;
width: 1900px;
height: 1200px;
position: relative;
}
The reason relative is used, is because it doesn't take the element out of the document flow (like fixed or absolute would) and as long as you don't specify a top, left, 'bottom', or right attribute to the parent (#background in the case), it will stay in the same location as it would with default positioning.
Edit:
I don't think this will work out of the box for you. You need to figure out how to make the image's width dynamic as well. You can either give it a % based width or use media queries.
Edit 2:
Ia also just noticed you have position:fixed for img#element. Change that to position:absolute. that will make it so that it is positioned relative to the position:relative parent rather than the window.
Consider making a javascript function that calculates the screen width. After that add margin-left to #background equal to ( screen width / -2 ). Make #background width & height - 100%
The div is 50% opaque and is displayed on top of the site's content, and it has a fixed width of 960 pixels (it's created by jQuery).
How can I center it horizontally?
margin: 0 auto; doesn't work :(
margin: 0 auto; won't work for elements that are absolutely positioned. Instead, we can work with the properties of the element itself. Check out the fiddle...
http://jsfiddle.net/UnsungHero97/HnzyT/2/
To center it horizontally is easy since you know the width AND its positioned absolutely. All you have to do is give it 2 CSS properties:
width: 960px;
position: absolute;
/* ... other CSS properties... */
left: 50%; /* 1. move it to the right 50% of the width of the page; now the left side of the element is exactly in the middle */
margin-left: -480px; /* move the element to the left exactly half of its width and now the center of the element is centered horizontally */
If you want to center it both vertically and horizontally, you need to know how wide AND how tall the element is.
I hope this helps.
Hristo
Just subtract the window midpoint from half the width of you element on resize. Here's a simple plugin that you could easily accomodate to center vertically if need be as well:
$.fn.centerMe = function () {
this.css('left', $(window).width()/2 - $(this).width()/2);
};
$(window).resize(function() { $('#yourElem').centerMe(); });
$('#yourElem').centerMe();
See working example →
Is it possible to place 1 DIV inside another DIV and have the DIV inside have a larger width and height than the DIV it is contained within?
Sounds like a riddle....
Basically I want to have a container that the user can set the width and height of. They will have also selected and image, cropped it and resized it using jQuery. The image will be the background of the container, however due to the fact that the background image could be made bigger than the container I want it to be possible for the background to be a DIV that can expand beyond the height and width of its container - To crop the image if you like.
Do able?
Yes, if I'm understanding you correctly
the container would be relatively positioned, the div "inside" would be absolutely positioned inside it -
the absolute positioning co-ordinates and getting the image centered would be done something like this, at default,
#inner {
position: absolute;
top: 0 ;
left: 0;
right: 0;
bottom: 0;
background: url(theimage.jpg) no-repeat 50% 50%;
}
this should center the image in the user sized container
then the "cropping tool" would be able to manipulate the co-ordinates (in an equal measure I presume) either + or - those 0 values, -, negative, ones will allow it to expand outside the "outer" container
$('<div>').attr('id', 'innerdiv').appendTo($('#div'));
CSS
#div{
height: 100px;
width: 100px;
background: green;
}
#innerdiv{
height: 200px;
width: 200px;
background: red;
}
HTML
<div id="div"></div>
http://jsfiddle.net/bKetM/
A background image will be 'cropped' by default. For example, if I have a 500px by 500px div and then I put a 1000px by 1000px image as its background then it will show the top left 500px by 500px of that image as a background. I could set the background image to be centered like so:
background:transparent url(image.png) no-repeat center;