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JQuery Hide/Show on Scroll down

I'm new to jquery. I'm trying to write a script that will hide the div "box" and all children. When the user scrolls to the bottom of the page, the div "box" and all children display. For time's sake, we'll say the children are "chamber1", "chamber2" and "chamber 3".
when I hide "box", it only removes that div.
$(document).ready(function() {
$("#box").hide();
});
Apologies for lack of code, but I'm having trouble understanding this lesson and I can't find an exact example of what I'm trying to do through my internet searches.
Thank you!

If you to hide the box when you reach the bottom of the page, you javascript should be as follows:
JAVASCRIPT:
$(document).ready(function() {
$(document).on("scroll", function(){
if ( window.scrollMaxY == window.scrollY ) {
$("#box").hide();
}
})
});
HTML:
<div id="box">
<div>Chamber 1</div>
<div>Chamber 2</div>
<div>Chamber 3</div>
</div>

You should make sure that the div has id "box". If you're working with a div of class "box" then you would use:
$(document).ready(function() {
$(".box").hide();
});

I think this might help you and would be better to understand. A good explantion is given below here with demo.
$(window).scroll(function() {
if ($(window).scrollTop() + $(window).outerHeight() == $(document).outerHeight()) {
//Script for activity on reaching bottom of document
$("#box").fadeOut();
} else // optional
{
$("#box").fadeIn();
}
});
body {
margin: 0px;
width: 100%;
}
.container {
position: relative;
height: 900px;
width: 100%;
background: #fee;
}
#box {
position: fixed;
top: 50px;
left: 0px;
background: lightblue;
height: auto;
padding: 15px;
overflow: hidden;
max-width: 250px;
width: 210px;
}
#box > div {
margin: 5px;
background: #F33636;
padding: 10px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<div class="container">
</div>
<div id="box">
Box
<hr>
<div class="chamber1">
Chamber 1
</div>
<div class="chamber2">
Chamber 2
</div>
</div>
JSFiddle
You can play around with Fiddle Link.

show loading div with overlay on submit while waiting

My divs are not showing when I click on submit.
I can get them to show if I do a window.onload() but the divs have to have display: none; by default;
How can I make it so these divs show when I hit submit because my form takes about 30 seconds to process, it has a lot of fields.
HTML
<div id="overlay-back"></div>
<div id="overlay">
<div id="dvLoading">
<p>Please wait<br>while we are loading...</p>
<img id="loading-image" src="img/ajax-loader.gif" alt="Loading..." />
</div>
</div>
Submit Button
<div class="form-buttons-wrapper">
<button id="submit" name="submit" type="submit" class="form-submit-button">
Submit
</button>
</div>
CSS
#overlay {
position : absolute;
top : 0;
left : 0;
width : 100%;
height : 100%;
z-index : 995;
display : none;
}
#overlay-back {
position : absolute;
top : 0;
left : 0;
width : 100%;
height : 100%;
background : #000;
opacity : 0.6;
filter : alpha(opacity=60);
z-index : 990;
display : none;
}
#dvLoading {
padding: 20px;
background-color: #fff;
border-radius: 10px;
height: 150px;
width: 250px;
position: fixed;
z-index: 1000;
left: 50%;
top: 50%;
margin: -125px 0 0 -125px;
text-align: center;
display: none;
}
jQuery
<script>
$(function() {
$('#submit').on('submit', function() {
$('#dvLoading, #overlay, #overlay-back').prop("display", "block").fadeIn(500);
});
});
</script>
The reason I am displaying none by default in css because if someone has javascript disabled I do not want any inteference

Please provide your own custom form validation as I have no context to supplement that. This should be placed in a document ready OR in a setInterval JavaScript function (the latter typically yeilds much better results).
$('button#submit').click(function() {
If (formValid === true && $('#dvLoading, #overlay, #overlay-back').not(':visible');)
{
$('#dvLoading, #overlay, #overlay-back').toggle(500);
$('button#submit').toggle(500); //hide this to prevent multiple clicks and odd behavior
} else {
var doNothing = "";
}
});

Try this:
<script>
$(function() {
$('#submit').on('click', function(evt) {
evt.preventDefault();
$('#dvLoading, #overlay, #overlay-back').fadeIn(500);
});
});
</script>
The fadeIn will make the div visible.
.prop sets the properties of the element, not the style.
Changed to use the click event

How to position two elements centered on top of each other?

The problem:
I have a form with a button underneath it to submit (post) from data with jQuery ajax(). I want for the button to be replaced with a spinner (animated png) for the duration of server ajax call. But such a trivial task is impossible in css to do right.
What i have tried:
I have placed button and image inside a bootstrap row. Ox ajax call I have set button display to none and img display to block. But because this two are not of the same size makes the whole page flicker, breaks the positioning of other elements and so on.
Another idea was to try to place both elements on top of each other with absolute positioning. But, stupid as css is I cannot center it on the middle of the row.
Is there a way to position both elements on top of each other so I can control their visibility?
Please bear in mind that I cannot used absolute position in pixel, because this is a web page and I do not not how wide the browser will be, image can change in the future, text in the button can change in the future, all this things affect absolute size.
If there is another solution to my problem which would prevent the page from jumping up and down it would also be great.
EDIT
Link to one of fiddle experiments:
https://jsfiddle.net/ofb2qdt8/
.button {
position: relative;
margin: auto;
height: 50px;
width: 30px;
background: blue;
z-index: 1;
display: block;
}
.spinner {
position: relative;
margin: auto;
height: 30px;
width: 50px;
background:red;
z-index: 2;
}
This renders second element underneath on screen. Not on different z layer.
Experiment 2:
https://jsfiddle.net/ofb2qdt8/
.button {
position: absolute;
margin: auto;
height: 50px;
width: 30px;
background: blue;
z-index: 1;
display: block;
}
.spinner {
position: absolute;
margin: auto;
height: 30px;
width: 50px;
background:red;
z-index: 2;
}
This does not center both elements, and they are pushed to the top of the containing div. The element with less height should be centered.

Check this working demo: https://jsfiddle.net/ofb2qdt8/3/
Add in a few lines of jquery and update your css.
Position your loading div according to button div's position, width, height using jquery.
*Click the button to see loading div, and try to play the margin of the button to any pixel.
###JQUERY
$(document).ready(function () {
$('.c2').each(function () {
$(this).css({
'width': $(this).siblings('.c1').outerWidth(),
'height': $(this).siblings('.c1').outerHeight(),
'top': $(this).siblings('.c1').offset().top,
'left': $(this).siblings('.c1').offset().left
});
});
$('.c2').on('click', function () {
$(this).hide(0);
});
});
###CSS
.c1 {
margin: 100px auto;
width: 100px;
text-align: center;
padding: 5px 10px;
background: blue;
z-index: 1;
}
.c2 {
position: fixed;
text-align: center;
background: red;
z-index: 2;
cursor: pointer;
}

Rough, ready and untested:
HTML
<div>
<input type='submit' />
<img src="spinneyIMage.gif" />
</div>
CSS
div{ text-align: center; }
div img{ display: none; }
jQuery
$('submit').click(function(e){
e.preventDefault();
$(this).hide().next().show();
});
After the Ajax call completes reverse the above jQuery.

As I haven't been able to find a working solution I have reverted to my first idea which I discarded at first. Albeit with a little twist.
HTML
<div class="row>
<div id="container-button" class="col-xs-12>
<button id="button" onclick="button_OnClick(e)">submit form via ajax</button>
<img src="img/spinner.png" sytle="display: none" />
</div>
</div>
JS
function btnContact_OnClick() {
// show the soinner and hide the button
showSpinner();
$.ajax(
{
type: 'POST',
url: "someurl.com/target",
data: $("#form").serialize(),
dataType: "json",
complete: function() { hideSpinner();},
success: onAjaxSuccess,
error : onAjaxError
});
}
function hideSpinner() {
$("#spinner").hide();
$("#button").show();
// make container height non-fixed and content adjustable again
$("#container-button").height('auto');
}
function showSpinner() {
// THis is the trick !!!
// Make the container fixed height as it was before displaying spinner, so it does not change with content (spinner is not the same height as button
$("#container-button").height($("#container-button").height());
$("#button").hide();
$("#spinner").show();
}
This is not the perfect solution but the best I could make.
Drawbacks:
it is not clean, you have to use javasript to fix what is css layout
problem
it still causes a little flicker
the height of container while displaying spinner is dependant on button, this may cause clipping if spinner is too big

Horizontal slide banner design & functionality

I need to show the vertical banner which will slide out horizontally from the right of the screen once user click the open button/div and should be able to close the same. Base design is show in the sample image below
I have set up fiddle for the same http://codepen.io/anon/pen/oXgePX
<div class="horizontal-slide" id="other-menu"><<</div>
<div class="menu-other slide-right slide-opened" id="other-menu-content">
horizontal on right slide div
</div>
UPDATE:
I managed to do it by wrapping banner & button inside another div.
.wrapper{
float:right;
position:absolute;
right:0;
}
example: http://codepen.io/anon/pen/aOzyxo
still need to improve it so that button also slides with the banner for smooth look

You've to change your css
For example:
/* float */
float:right;
/* absolute position */
position:absolute;
You need it like this:
http://codepen.io/anon/pen/mJyMgW
Or do I missunderstand your Question?

Maybe you wanted something like this. It's a good start I think.
$(function() {
//slideout-menu-toggle
$(".banner-slide").click(function(event) {
var _slideMenu = $("#right-banner");
if (_slideMenu.hasClass("slide-opened")) {
$( "#right-banner" ).animate({ "right": "-190px" }, "slow" );
$( "#hbanner" ).html("<<");
//alert('a')
} else {
$( "#right-banner" ).animate({ "right": "0" }, "slow" );
$( "#hbanner" ).html(">>");
}
_slideMenu.toggleClass("slide-opened");
});
});
#right-banner {
position:absolute;
right:0;
top: 0;
width: 210px;
}
.right-banner {
width: 150px;
height: 200px;
background-color: green;
padding: 20px;
float: right;
background-color: blue;
}
#hbanner {
float: left;
background-color: red;
cursor:pointer;
}
.wrapper {
float: right;
position: absolute;
right: 0;
z-index: 100000;
}
.content {
width: calc(100% - 210px);
background-color: magenta;
height: 200px;
float: left;
text-align: right;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<div class="content">
This is sample text</div>
<div class="wrapper">
<div id="right-banner" class="slide-opened">
<div class="banner-slide" id="hbanner">
>>
</div>
<div class="right-banner" id="hbanner-wrapper">
horizontal on right slide div
</div>
</div>
</div>

You can do it cleaner and simpler using CSS transitions and a simple toggleClass in jQuery:
$('#button').on('click', function(){
$('#banner').toggleClass('unfolded');
});
Demo : http://codepen.io/mbrillaud/pen/NqPvZM

use a JQ toggle class
$('#slider').toggleClass('sliderthing');
that plus CSS to give it looks and stuff should work well

I managed to do it by wrapping the banner & button inside another wrapper.
Here is the updated example : http://codepen.io/anon/pen/aOzyxo
I have added some exact animation so that red button hide/shows immediately upon click event with some smooth animation.
.wrapper{
float:right;
position:absolute;
right:0;
z-index:100000;
}

Onclick change top position doesnt do anything

I think the problem's already stated. When you click the plus extra content is show(the Hello Guest and Register and Signup). Is their a better and working way of doing this or do I have a bug. New at java-script so don't hate.
---> http://jsfiddle.net/CM9Av/

You are not setting the top . Also, cache your variables.
function menuanimate() {
var loginoutbox = document.getElementById("loginoutbox");
var fromtop = loginoutbox.style.top;
if (fromtop == "-20px") {
fromtop = "0px";
}
else {
fromtop = "-20px";
}
loginoutbox.style.top = fromtop;
}​
In markup , set top in inline style. stlye.top wont read css property.
<div id="loginoutbox" class="ablack" style='top:-20px'></div>
Try outside fiddle in a simple webpage. It works !
Demo fiddle (I dont know how to put pure js in fiddle.net)

You can not just update the variable fromtop, that will have no effect on the page. You will have to update the top-value manually, the same way you read it.

<html>
<script type="text/javascript">
function menuanimate(){
document.getElementById("loginoutbox").style.top="0px";
}
</script>
<style type="text/css">
#image{
position:fixed;
top:0px;
left:80%;
}
#image:hover{cursor:pointer;}
#loginoutbox{
left: 80%;
width: 218px;
color: white !important;
position: fixed;
top: -20px;
background-color: #444;
height: 20px;
text-align: right;
}
</style>
<img style="margin-right: 94px;" src="http://www.kdogisthebest.webege.com/images/plustab.png" id="image" onClick="menuanimate();"/>
<div id="loginoutbox" class="ablack" style="top:-20px;">
<div style="display: inline-block; font-size: 14px; padding-left: 20px;">Hello Guest!</div>
<a id="register" href="#fallr-register" class="button">Register</a> |
<a id="signin" href="#fallr-signin" class="button">Login</a>
</div>
</html>
This is not the prettiest but I think this is what you are looking for as far as the end result...