This could be really obvious and I'm completely missing it.
I've searched for hours and can't seem to find a way to, using jQuery, reveal a hidden div from the bottom up. What I am trying to achieve is exactly as in the following link, but in reverse: http://jqueryui.com/demos/show/
I can slide a div from the bottom to the top, but this reveals itself as it moves, rather than being 'masked' in.
Like I said, this could (should?) be really obvious and I'm not seeing it, but I've been looking for ages and can't find a solution to this relatively simple problem.
Thanks,
Ronnie
The effect you're looking for is a little tricky to achieve, but it can be done, even without a wrapper element.
The main issue here is that elements naturally render top-down, not bottom-up. Animating both the top
and height
CSS properties of a relatively-positioned element allows us to implement a slide-up effect, but the element will still render top-down:
+-------------------------------------------+
| | ^
| | | Hidden area collapses upwards.
| | |
+-------------------------------------------+ <-- 'top'
| | ^
| | Upper part of element (visible). | |
| | | | Animation goes bottom-up.
| | Element still renders top-down. | |
| | | |
+--|----------------------------------------+ <-- 'top + height'
| | | |
| | Lower part of element (hidden). | |
| V | |
+-------------------------------------------+
If we want to simulate bottom-up rendering, we have to modify the scrollTop property of the element during the animation, in order for its lower part to always remain in view:
+-------------------------------------------+
| | ^
| | Upper part of element (hidden). | | Hidden area collapses upwards.
| | | |
+--|----------------------------------------+ <-- 'top' and 'scrollTop'
| | | ^
| | Element still renders top-down. | |
| | | | Animation goes bottom-up.
| | Lower part of element (visible). | |
| V | |
+-------------------------------------------+ <-- 'top + height'
We can use animate() with scrollTop
, but doing so in conjunction with top
and height
did not work correctly in my tests (I suspect scrollTop
is reset when top
or height
are modified in the first animation step, so it ends up stuck to 0
).
To work around this, we can handle scrollTop
ourselves through the optional step function we can pass to animate()
. This function is called with two arguments, now
and fx
, now
being the current value of the animated property and fx
being a wrapper object around useful information, like the element and property being animated.
Since we always want scrollTop
to be the same as top
, we only have to test if top
is being animated in our step
function. If it is, we set scrollTop
to now
. This solution gives acceptable results, although it flickers a little too much for my taste (that might be an artifact of my browser, though).
So, in summary, to implement that effect, we have to:
position: relative;
so we can animate its top
property,top
to the original height and height
to 0
,display: none;
is applied),top
to 0
and height
to the original height,scrollTop
the value of top
on each animation step.Resulting in the following code:
$("#click").click(function() {
var $revealMe = $("#revealMe");
var originalHeight = $revealMe.height();
$revealMe.css({
position: "relative",
top: originalHeight,
height: 0
}).show().animate({
top: 0,
height: originalHeight
}, {
duration: 1000,
step: function(now, fx) {
if (fx.prop == "top") {
$(fx.elem).scrollTop(now);
}
}
});
});
You can test it in this fiddle.
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