"Zooming" elements on a page while keeping the centre of enlargement in the centre of the window
I'm trying to work out how to enlarge all elements on a page, but keep the centre of enlargement in the centre of the window.
On this page, once the image reaches the top or the left side of the window开发者_如何学Go the centre of enlargement changes. It also changes when you move the image. (exactly what you would expect)
I'm thinking I'd need to take a completely different approach to achieve what I want. But I'm not sure what that approach is..
Any ideas?
Well, here's my take.
Only thing is that I ditched the containers you were using. Is that cheating? Seems like they were only there to get the image centered. No need.
This works as expected with no side effects.
Here's a working demo you can test:
http://jsfiddle.net/YFPRB/1/
(You need to click on the pane with the baboon first.)
HTML
<body>
<img src="http://cdn.sstatic.net/stackoverflow/img/apple-touch-icon.png" />
</body>
CSS
html, body {
width: 100%;
height: 100%;
overflow: hidden;
}
jQuery
EDIT: Thanks to @stagas for the reminder to clean up redundancies.
var $img = $('img'); // Cache the image. Better for performance.
$img.draggable();
$img.css({left: ($('body').width() / 2) - ($img.width() / 2)})
.css({top: ($('body').height() / 2) - ($img.height() / 2)})
$(document).keydown(function(event) {
if (event.keyCode == 38) {
var adjustment = 1.25;
} else if (event.keyCode == 40) {
var adjustment = 0.8;
} else {
return;
}
var offset = $img.offset();
var width = $img.width();
var height = $img.height();
var newWidth = width * adjustment;
var newHeight = height * adjustment;
var diffWidth = newWidth - width;
var diffHeight = newHeight - height;
var hcenter = $('body').width() / 2;
var vcenter = $('body').height() / 2;
var leftPercent = (hcenter - offset.left) / width;
var topPercent = (vcenter - offset.top) / height;
$img.offset({top: offset.top - (diffHeight * topPercent), left: offset.left - (diffWidth * leftPercent)});
$img.width(newWidth).height(newHeight);
});
This is what I came up, it works as you say except the image will always go to the center after zooming in or out:
$('document').ready(function() {
zoomimg=$('#zoomimg'); // we store this in a variable since we don't need to traverse the DOM every time -- this is faster
var viewportWidth = $(window).width();
var viewportHeight = window.innerHeight ? window.innerHeight : $(window).height(); // this is to work with Opera
zoomimg.css({'position': 'absolute', 'left': (viewportWidth/2)-(zoomimg.width()/2), 'top' : (viewportHeight/2)-(zoomimg.height()/2)}).draggable();
$(document).keydown(function(event) {
event = event || window.event;
var viewportWidth = $(window).width();
var viewportHeight = window.innerHeight ? window.innerHeight : $(window).height(); // this is to work with Opera
if (event.keyCode == 38) {
width = zoomimg.width();
height = zoomimg.height();
zoomimg.width(width*1.2).height(height*1.2);
var viewportWidth = $(window).width();
var viewportHeight = window.innerHeight ? window.innerHeight : $(window).height();
zoomimg.css({'left': (viewportWidth/2)-(zoomimg.width()/2), 'top' : (viewportHeight/2)-(zoomimg.height()/2)});
} else if (event.keyCode == 40) {
width = zoomimg.width();
height = zoomimg.height();
zoomimg.width(width*0.8).height(height*0.8);
var viewportWidth = $(window).width();
var viewportHeight = window.innerHeight ? window.innerHeight : $(window).height();
zoomimg.css({'left': (viewportWidth/2)-(zoomimg.width()/2), 'top' : (viewportHeight/2)-(zoomimg.height()/2)});
} else {
return
}
});
});
You should put an ID 'zoomimg' on the tag for it to work, and overflow:hidden on the #container . Also ditch that display:table and display:table-cell they're useless now that we center with Javascript. Also, pressing the down arrow key will cause the container to scroll down, so you should use other keys, as the arrows are reserved by the browser for scrolling the viewport.
加载中,请稍侯......
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