Controlling slick slider with external links - javascript

I am using slick slider. I have created two div's. Here is details :)
This is the css for these two divs
.Full_Left {
position: absolute;
top: 0px;
width: 50%;
height: 100%;
background: rgba(34, 25, 165, 0.5);
left: 0px;
z-index: 2;
}
.Full_Left:hover {
cursor: url(../images/right.svg), auto;
}
.Full_Right {
position: absolute;
top: 0px;
width: 50%;
height: 100%;
background: rgba(216, 33, 33, 0.5);
right: 0px;
z-index: 2;
}
.Full_Right:hover {
cursor: url(../images/left.svg), auto;
}
as you can see now my div's are on full screen and I want to control the slider with the links I have created so that when user bring mouse on 50% he have right arrow and same for left side and all of the slider content will be below these two layers.
I am wondering this is possible with "Slick" or not ?
Thanks in advance.

I had a similar use case to this, though with a twist. Finding this question pointed me in the right direction, and my answer below does address controlling a Slick slider with external links.
I have a Slick slider in the right column set to position: fixed and a longer story in the right column that scrolls. Each of the slides in the slider corresponds to a section of the story, set apart by <h3> tags. I wanted a link next to each header that set the slideshow to the corresponding slide when clicked. To accomplish this, I added a button at the end of each section header:
<h3>Section Header <button data-slick-index="0" class="slide-to"><i class="fa fa-camera-retro" aria-hidden="true"></i></button></h3>
The data-slick-index property (is it a property?) was set to the slick-index of the corresponding slide. Then just below my Slick init JS I placed the following:
$(".slide-to").bind("click", function() {
var slidx = $(this).attr('data-slick-index');
$('#slideshow').slick('slickGoTo', slidx);
});
Works like a charmp*.
That's a charm and a champ combined.

I have solved this problem by my self. here is my working code
<script type="text/javascript">
$('.Full_Right').click(function(e) {
$('#Contrroll').slick('slickNext');
});
$('.Full_Left').click(function(e) {
$('#Contrroll').slick('slickPrev');
});
</script>
:)

Related

Page doesn't scroll when cursor is over fixed div

My layout looks almost identical to this codepen.
.parent {
color: white;
padding: 70px;
position: relative;
background-color: #0074d9;
margin-top: 50px;
}
.element {
background-color: lighten(#0074d9, 20);
opacity: .85;
padding: 20px;
color: rgba(255,255,255,.9);
position: fixed;
bottom: 0;
left: 0;
right: 0;
}
The codepen works the right way, so it's been hard to come up with a demonstrable example.
When my cursor is positioned over the fixed "child element" div, I want to be able to scroll the parent but not be able to clickthrough.
The common answer seems to be "pointer-events: none", but that allows click interaction with the page below.
Open to other suggestions or explanations as to why it works in the codepen, but doesn't outside of it.
The solution that worked for me was to use jquery to grab the parent by id and add my deltaY to its scrollTop.
<div
onWheel={(e) => {
const component = $(`#content`);
const contentScrollPosition = component.scrollTop();
component.prop("scrollTop", contentScrollPosition + e.deltaY);
}}
</div>
This allows me to scroll the parent even when my cursor is on the fixed-position div.
Unfortunately, mobile doesn't work well. First, you would need to track the touch event though onTouchStart, End, and Move. Even then, you lose touch scroll momentum which makes it feel too unnatural.

Scrolling Nav Sticks to Top

My problem is along the lines of these previous issues on StackOverflow but with a slight difference.
Previous issues:
Stopping fixed position scrolling at a certain point?
Sticky subnav when scrolling past, breaks on resize
I have a sub nav that starts at a certain position in the page. When the page is scrolled the sub nav needs to stop 127px from the top. Most of the solutions I have found need you to specify the 'y' position of the sub nav first. The problem with this is that my sub nav will be starting from different positions on different pages.
This is the JS code i'm currently using. This works fine for one page but not all. Plus on mobile the values would be different again.
var num = 660; //number of pixels before modifying styles
$(window).bind('scroll', function () {
if ($(window).scrollTop() > num) {
$('.menu').addClass('fixed');
} else {
$('.menu').removeClass('fixed');
}
});
I'm looking for a solution that stops the sub nav 127px from the top no matter where on the page it started from.
You can use position: sticky and set the top of the sub-nav to 127px.
See example below:
body {
margin: 0;
}
.main-nav {
width: 100%;
height: 100px;
background-color: lime;
position: sticky;
top: 0;
}
.sub-nav {
position: sticky;
width: 100%;
height: 50px;
background-color: red;
top: 100px;
}
.contents {
width: 100%;
height: 100vh;
background-color: black;
color: white;
}
.contents p {
margin: 0;
}
<nav class="main-nav">Main-nav</nav>
<div class="contents">
<p>Contents</p>
</div>
<nav class="sub-nav">Sub-nav</nav>
<div class="contents">
<p>More contents</p>
</div>
Please see browser support for sticky here
You should change your code to the below, should work fine:
$(window).bind('scroll', function () {
if ($(window).scrollTop() > $(".menu").offset().top) {
$('.menu').addClass('fixed');
} else {
$('.menu').removeClass('fixed');
}
});
Maybe you can try this:
Find navigation div (.menu)
Find the top value of the .menu (vanilla JS would be menuVar.getBoundingClientRect().top, not sure how jQuery does this).
Get top value of browserscreen.
Calculate the difference - 127px.
When the user scrolls and reaches the top value of the menu -127px -> addClass('fixed').

How to create an interactive navigation, in a circular design, using HTML, CSS, and JQuery

// JavaScript Document
$('.page').hide();
$(".btns").click(function(e){
e.preventDefault(); //this method stops the default action of an element from happening.
var $me = $(this); //$(this) references .btns, the object in local scope.
var $myContent = $($me.attr('href')); //pulls href of page 01, 02, or 03.
$('.page').hide(); //hides all pages
$myContent.fadeIn();//fades in clicked href connected to btn
$(".btns").removeClass('selected');//
$me.addClass('selected');
});
*{
border-spacing: 0px;
}
body{
margin: 0px;
padding: 0px;
}
.circle-container {
position: relative;
width: 24em;
height: 24em;
padding: 2.8em;
/*2.8em = 2em*1.4 (2em = half the width of a link with img, 1.4 = sqrt(2))*/
border: dashed 1px;
border-radius: 50%;
margin: 1.75em auto 0;
}
.circle-container a {
display: block;
position: absolute;
top: 50%; left: 50%;
width: 4em; height: 4em;
margin: -2em;
}
.circle-container img { display: block; width: 100%; }
.deg0 { transform: translate(12em); }
<div class="body_content">
<div class="page" id="page_01">
<h2>1. Category 1</h2>
</div>
</div>
<div class="circle-container">
<nav class="navigation">
<a href="#page_01" class="btns deg0" >
<img id="one" src="imgs/button.png" alt="page01"/>
</a>
</nav>
</div>
I have a unique situation that I would like to discuss with you all. I am trying to create a web page that has a circular navigation, as shown here enter image description here
Each one of these buttons would display content when clicked, like an in-page link. The JQuery is as shown enter image description here
The concept seems simple enough, force all content to hide, when a user clicks a button, the page content linked to that button shows. It works when the links are inline or block display, but when in a circle, the links don't work, the button content doesn't show. Has anyone worked with a similar issue? Or would anyone have a potential solution? I apologize for the vagueness of the questions but the issue seems multi-faceted. Any advice or ideas would be greatly appreciated.
Are you sure your jQuery reference is working? I don't see any issue with the code, the click event should fire when you click on the links. Check the console for any errors, I strongly believe jQuery might not get loaded.

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

Place animated footer under other divs

so I wanted an animated footer for my webpage using jquery. There's supposed to be a button which should trigger the animation. I found a nice example for all this, and everything is fine and dandy. Except that the button (including the footer) has this code that makes it stick to the bottom of your web browser, rather than to the bottom of the page. I do [i]not[/i] want it to, like, "scroll" along with the page, I realy want it to be underneath all my other divs. I tried putting it in the div container (which has all my other divs in it as well), but that doesn't seem to work.
Now, (after 2.5 hours of googling) I found out that it might/may/could have something to do with "absolute" positioning in the CSS, so I tried switching some things around such as giving the footer's container a relative position or giving it an "overflow: hidden;" along with the rest a left float but nothing seemed to solve my problem. (I could've done something wrong, not that great with CSS after all :-/)
I hope someone is able/willing to help.
P.S. Here's the example I used:
http://return-true.com/2010/04/jquery-pop-up-footer-version-2/
and here's the code:
Javascript:
jQuery(function($) {
var open = false;
$('#footerSlideButton').click(function () {
if(open === false) {
$('#footerSlideContent').animate({ height: '300px' });
$(this).css('backgroundPosition', 'bottom left');
open = true;
} else {
$('#footerSlideContent').animate({ height: '0px' });
$(this).css('backgroundPosition', 'top left');
open = false;
}
});
});
HTML:
<div id="footerPlacement">
<div id="footerSlideContainer">
<div id="footerSlideButton"></div>
<div id="footerSlideContent">
<div id="footerSlideText">
<h3>Hey! I'm a Sliding Footer</h3>
<p>What's a Sliding Footer? Well I'm a cool little element which can be hidden from view, and revealed when the user wants to see me.</p>
<p>What can you use me for? Well look at all this stuff:</p>
<ul>
<li>Sales information</li>
<li>Important updates</li>
<li>Unobtrusive about panel</li>
<li>Or just a good ol' footer</li>
</ul>
<p>There are obviously many other uses, but these are the few useful ones I can think of.</p>
</div>
</div>
</div>
</div>
CSS:
#footerPlacement {
margin-bottom: 0px;
width: 1000px;
margin-left: auto;
margin-right: auto;
}
#footerSlideContainer {
position: fixed;
margin-left: 0px;
bottom:0px;
width: 1000px;
}
#footerSlideButton {
background: url('../images/footer/footerbtn.png') top left no-repeat transparent;
position: absolute;
top: -55px;
right: 20px;
width:50px;
height:50px;
border: none;
cursor: pointer;
}
#footerSlideContent {
width: 100%;
height: 10px;
background: #251b15;
color: #CCCCCC;
font-size: 0.8em;
border: none;
font-family: DejaVuSansBook, Sans-Serif;
}
#footerSlideText {
padding: 15px 10px 25px 25px;
}
Thanks in advance!
if you change your #footerPlacement to include position:relative, you can change #footerSlideContainer to be position:absolute and then your footer will sit below any content above it.
However you will need to make the content have a min-height of around 350px for the footer to work properly and if your content isn't long enough, the footer won't be at the bottom of the browser.
I also added overflow:hidden to #footerSlideContent. I have made a fiddle to demonstrate:
http://jsfiddle.net/tc6b8/

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