We Keep Coding

How to make an element appear then disappear on click

I have been trying to make a random image appear on click by adding a fadeOut effect and then removing the class. when I click it works fine, but I don't know how to remove the class after a few milliseconds and then being able to appear again on another click. so far I have just been able to make it fade out on click, I have tried a setInterval function so that the class gets removed after 1 millisecond but didn't work so I erased it, but even then, I don't know how to make the .on('click', function()) function fire on every click, instead of just working once. any help or tips would be really appreciated. Thanks!
<style>
body {
background-color: black;
}
img {
opacity: 0;
width: 40px;
z-index: 0;
position: relative;
top: 3em;
}
</style>
<img class="red"
src="http://www.clker.com/cliparts/0/f/1/f/130267960774173786paint-
splash(red)-md.png" alt="">
<img class="blue" src="http://www.clker.com/cliparts/Q/3/H/u/Z/K/dark-blue-
splash-ink-hi.png" alt="">
<img class="yellow" src="http://www.clker.com/cliparts/3/y/m/m/p/P/yellow-
splash-ink-md.png" alt="">
<script>
$(document).ready(function(){
var red = $(".red");
var blue = $(".blue");
var yellow = $(".yellow");
var images = [red, blue, yellow];
$(document).on('click', function(){
$(images[(Math.floor(Math.random()*3))]).addClass("animated fadeOut");
});
})
//i should be able to click anywhere on the screen and a random image should appear and then fadeout each time there is a click
</script>

Try something like this:
$(document).on("click", function() {
$("#element").show(0, function() {
$("#element").fadeOut();
});
});
$("#element").hide();
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<span id="element">Element</span>

It looks like you are using jQuery so you simply need to:
1) Create a function that hides the class. Example:
function hideStuff(){
$(".myimg").hide();
}
2) Add a class to your image files so they have a common selector (like "myimg" below). You may also want to add an "alt" attribute (was missing in your code). Example:
<img class="yellow myimg" src="http://www.clker.com/stuff" alt="image-three">
3) Add the timeout as part of your function with the amount of delay you want. While it is not required, you should include a variable name so you can call it in the future. Example:
var myTimeout = setTimeout( hideStuff, 5000);
Hopefully these will get you going in the right direction.

Both .fadeOut() and .hide() set display: none, which could effect your layout. I think you're looking to animate opacity to 0, and then in the callback function you can change the image source. I'd recommend using a div and setting the background-image property since divs are a bit more layout friendly. Also, you could either use classes and set the background-image property in the <style> section or you can make an array of the image urls and randomly pick from that (which is what I did here).
let images = [
'http://www.clker.com/cliparts/0/f/1/f/130267960774173786paint-splash(red)-md.png',
'http://www.clker.com/cliparts/Q/3/H/u/Z/K/dark-blue-splash-ink-hi.png',
'http://www.clker.com/cliparts/3/y/m/m/p/P/yellow-splash-ink-md.png'
];
$(document).on('click', function() {
let $img = $('.img'); //so you don't have to make a new object everytime it's used
if ($img.css('opacity') === '1') {
$img.animate({ opacity: 0 }, function() {
$img.css('background-image', `url(${images[Math.floor(Math.random()*3)]})`);
});
} else {
$img.animate({ opacity: 1 });
}
}).click().click(); //two clicks to initialize image
https://jsfiddle.net/yc4e4nxb/3/
NOTE: JSfiddle doesn't seem to like wherever these images are hosted, so it's working kind of erratically. Hopefully you get the gist of what this code is doing though.
http://api.jquery.com/animate/

If I understood the question correct, In This Fiddle the button element disappears when you click anywhere in the screen and then re appears immediately. Hope this will work.
$(document).ready(function(){
$(document).on('click',function(){
$("#myElement").fadeOut().delay(100).fadeIn();
});
});

Can't get hidden image to display with .show()

I'm using the vimeo api to slide a video off the screen after it finishes playing. Underneath the video player, hidden, I have an image that says 'replay'. The image though is slightly bigger than the player so I want to hide the image via .hide() or display: none in the css and then show it after the animation of the video player completes.
Here's my js:
$(document).ready(function() {
$(".vimeo-container img").hide();
$('iframe.vimeo').each(function(){
Froogaloop(this).addEvent('ready', ready);
});
function ready(playerID){
Froogaloop(playerID).addEvent('finish', onFinish);
}
function onFinish(playerID) {
var player = "#" + playerID;
$(player).animate({width: "0%"}, 750, function() {
$(player).next("img").show();
});
}
});
So the first line is hiding the image. And then when the onFinish function completes I'm trying to show() the image, but it won't work. I should note that when I reverse it and do:
$(player).next("img").hide();
it works.
Here's my HTML:
%section#container1
.row.video-left
.large-8.columns
.vimeo-container
.flex-video.widescreen.vimeo
%iframe.vimeo#player1{allowfullscreen: "", frameborder: "0", height: "225", mozallowfullscreen: "", src: "http://player.vimeo.com/video/60122989?api=1&player_id=player1", webkitallowfullscreen: "", width: "400"}
= image_tag "behind1.png", class: "behind1"
And CSS:
.vimeo-container {
position: relative;
.behind1 {
margin-top: -27em;
}
I've also tried setting display: none in the css, but that wont work either. Not sure what I'm missing.
Thanks.
EDIT
function onFinish(playerID) {
var player = "#" + playerID;
$(player).animate({width: "0%"}, 750, function() {
console.log($(player));
$(player).next().show();
});
}
When I log out ($(player) it returns:
And when I log out console.log($(player).next()); it logs out the image that I am trying to show.

According to the jQuery documentation on the animate method here:
Note: Unlike shorthand animation methods such as .slideDown() and .fadeIn(), the .animate() method does not make hidden elements visible as part of the effect. For example, given $( "someElement" ).hide().animate({height: "20px"}, 500), the animation will run, but the element will remain hidden.
I had a similar need in a project and what worked for me there was to set the z-index of the element I wanted to hide to be less than that of the background. Then, when I wanted to show (or, in your case, animate) I could apply the jQuery methods to the element as if they were hidden (by increasing the z-index so that the element becomes visible), yet not incur the undefined behaviour of attempting to manipulate a hidden element.
Another option would be to move the element off the screen by way of a negative (x, y) coordinate and work from there. I'm not sure which visually would be more appealing in your use case but mention it for completeness.

jQuery hover stuck

When I hover over an img which fades to another img and sroll off too fast, the fadeOut gets stuck and the fade stays. I've tried the .stop() as I've seen in other responses, but still won't work. Is there something else I can put instead of the .stop()?
<div class="grid big-square">
<a href="#"><img id="image2" src="img/fade/creo.png">
<img id="image1" src="img/creo.jpg"></a>
</div>
<script>
$("#image1").mouseenter(function () {
$(this).stop(true, true).fadeOut(1000);
});
$("#image2").mouseleave(function () {
$("#image1").stop(true, true).fadeIn(500);
});
</script>

I seem to remember having a similar problem when I was creating this website.
The solution is to use a combination of .hover() and .stop() to ensure that only one animation is running at a time, which I think you have. Also ensure that the mouseover image is on top of the other image, and just fade that one in and out. The image fading out gets 'stuck' because at some opacity the .mouseleave() stops firing and the .mouseenter() starts firing on the other image.
Something like:
$$ = $("#image1");
$$.hover(function () {
$$.stop().animate({
opacity: 0
}, 1000);
}, function () {
$$.stop().animate({
opacity: 1
}, 1000);
});
#image1 must be above #image2 for this to work, #image1 fades out to 'reveal' #image2 behind it. The code uses .animate() rather than .fadeIn() and .fadeOut() but the effect is the same.
Edit- to fade in another div after the end of the fadeoout animation use the complete call back of the animate function.
Something like:
$$ = $("#image1");
$$.hover(function () {
$$.stop().animate({
opacity: 0
}, 1000);
}, function () {
$$.stop().animate({
opacity: 1
}, 1000, function() {
$("#finalDiv").animate({ opacity: 1, 500 });
});
});
#finalDiv needs to be after the 2 <img />s in your html to appear above them.

I'm not sure how you're trying to accomplish this but I do know how it should be done.
Try this http://jsfiddle.net/xy5dj/
Make sure to listen for both events on the same element (preferably a wrapper element).
Take note that fadeOut actually removes the element from the rendered content (display: none) making sure that the mouse events won't fire on that element.
Side note:
A dirty trick that I used once (if you have to do this then you're doing something wrong) is to clear the style of the element after animation using the callback ability of the animate function i.e.
$('el').animate({opacity:0}, 500, function(){$(this).attr('style', '')});

fiddle
You should use the animation/transition in form:
.fadeTo( duration, opacity, complete )
where complete is callback function.

JavaScript How to make an image to change color onmouseover?

I have a button and an image and want them to change color onmouseover.
The button changes color fine:
<script>
function secondColor(x) {
x.style.color="#000000";
}
function firstColor(x) {
x.style.color="#ffaacc";
}
</script>
<input onmouseover="secondColor(this)" onmouseout="firstColor(this)" type="submit"><br>
How can I do the same thing with the image? Is there any way:
<img src="..." ......
Or do I have to have a second image to replace the first one onmouseover and this is the only way?

If you don't care that much about supporting older browsers, you could use the new CSS3 filter brightness. In chrome, you could write something like this:
var image = document.getElementById('img');
image.addEventListener('mouseover', function() {
image.setAttribute('style','-webkit-filter: brightness(1.5)');
}, false);
image.addEventListener('mouseout', function() {
image.setAttribute('style','-webkit-filter: brightness(1.0)');
}, false);
I don't recommend this approach, though. Using another picture while hovering would be a better solution.

I know that this is old, but you don't need two images. Checkout my example using one image.
You can simply change the position of the background image.
<div class="changeColor"> </div>
JavaScript
var dvChange = document.getElementsByClassName('changeColor');
dvChange[0].onmouseover = function(){
this.style.backgroundPosition = '-400px 0px';
}
dvChange[0].onmouseout = function(){
this.style.backgroundPosition = '0px 0px';
}
CSS
.changeColor{
background-image:url('http://www.upsequence.com/images/multibg.png');
width:400px;
height:400px;
background-position: 0px 0px;
}
.changeColor:hover{
background-image:url('http://www.upsequence.com/images/multibg.png');
width:400px;
height:400px;
background-position: -400px 0px;
}
You can also try changing the opacity of the images onmouseover and onmouseout.
I don't have an example for that, but its super easy to find and I am sure it has be answered already on stack exchange somewhere.
In the JSFiddle below there is Javascript and non-Javascript examples.
http://jsfiddle.net/hallmanbilly/gtf2s8ts/
Enjoy!!

I think you have to use a second image. I recently cam across the following article describing how to do image crossfading on hover using css. Crossfading Image Hover Effect

You can change image SRC on mouse over, you can load two images and use fade effects to "change" them. But better, you can use image as DIV background, make sprite and just move BG on mouse over.
Loading of two different images bring you to disappearing when hover and second image loading. Better do not use JS at all. Make sprite from two images, put it as BG of DIV and write two CSS for DIV, normal and when hover.

If you have access to JQuery use hover function. If you want to change image
$('#imageid').hover(function(){
//change image or color or opacity
$(this).attr('src', newImageSrc);
});
add this function in document ready function.

jQuery - Fading animation on hover

I have a little jQuery animation which fades in a link when hovering an :
$(function() {
$('.delete').hide();
$('#photos img').hover(function() {
$(this).parents('li').children('.delete').fadeIn('fast');
}, function() {
$(this).parents('li').children('.delete').fadeOut('fast');
});
});
But if I quickly move my mouse in and out of the image, the new animation is always added to the queue and when I stop I can see the link pulsating for a while. I tried using .stop(true), but then sometimes the fade in effect doesn't work at all (or just up to some opacity value less than 1). What can I do?
Thanks, Eric

The best way is to use hoverIntent plugin. This deals with the issues above. It also adds a slight delay to the animation so if a user happens to quickly move the mouse over all the links you do not get an ugly animation flow of all the links.

One way to prevent such problems occuring is to use stop() in conjunction with fadeTo(), as in the snippet below:
$(function() {
$('.delete').fadeTo(0, 0);
$('#photos img').hover(function() {
$(this).parents('li').children('.delete').stop().fadeTo('fast', 1);
}, function() {
$(this).parents('li').children('.delete').stop().fadeTo('fast', 0);
});
});
Hope this solves your issue!