I have a background image and lots of product image. Shortly, %88 of entire page is image. But when i refresh the page. Some products are displaying. After that my background is coming. Then other divs, images are coming.
What i want to do is my background images comes first. Then others are must be displayed. Sync or async it is not important.
What i tried:
i replaced my background image div to the top of the dom. Not worked.
window.onload=function () {
document.getElementById("theBackgroundImage").style.visibility = "visible";
}
Not worked also.
I tried several things before. But i have still same issue.
First of all, you will need to detect when all your images are loaded. Or preferably when all the images you want displayed first.
Then i see two options here. For both, i'm going to use a library that helps in detecting when images have been loaded : https://imagesloaded.desandro.com/
1st option
Just display your image.
imagesLoaded( 'img', function() {
document.querySelectorAll(".displayAfter").forEach(e => e.style.visibility = 'visible'
)
});
.displayAfter {
visibility : hidden;
}
<script src="https://unpkg.com/imagesloaded#4/imagesloaded.pkgd.min.js"></script>
Image here :
<img width="50px" height="50px" src="https://picsum.photos/50/50?random" />
Lazy loaded image here :
<img class="displayAfter" width="50px" height="50px" src="https://picsum.photos/50/50" />
2nd option
Do not load your image first (for instance src empty or pointing to blank image) then change the src attribute of your image. Here i use a data-src attribute in the image that i simply insert into the image src attribute once all the other images are loaded.
imagesLoaded( 'img', function() {
document.querySelectorAll(".displayAfter").forEach(e =>
e.setAttribute('src', e.getAttribute('data-src'))
)
});
<script src="https://unpkg.com/imagesloaded#4/imagesloaded.pkgd.min.js"></script>
Image here :
<img width="50px" height="50px" src="https://picsum.photos/50/50?random" />
Lazy loaded image here :
<img class="displayAfter" data-src="https://picsum.photos/50/50" width="50px" height="50px" />
Try to move the following to top in your css file
#theBackgroundImage{
background : url(..)
}
and maybe you need to use (Lazy Loading Images) => url
Related
When the user clicks a "Vacation" image on the left, a larger copy
of it will appear to the right of the images in the image with
an id of "currentimage" inside the "bigimage" div. (See the HTML file)
At the same time one of the following should appear below
the "currentimage": "Mountain Vacation", "Ocean Vacation", or "Desert Vacation"
depending on which image was selected.
this is part of the HTML code
<div id="bigimage">
<img id="currentimage" src="http://profperry.com/Classes20/JQuery/ocean.jpg" alt="ocean vacation" width="300" height="225" border="0">
<p id="imagedesc"></p>
</div>
This is part of the JS code
$("img").click(function ()
{
var mySrc = $(this).attr("src");
$("#currentimage").attr("src", mySrc);
$("#imagedesc").html(this.alt);
});
I also tried
$("img").click(function ()
{
$("#currentimage").attr("src", this.src);
$("#imagedesc").html(this.alt);
});
but when i click the image a larger copy of it is not appearing on the right
Browsers not reload image after change src. Try remove old element with old image and paste new element with new image src.
Or: How to reload/refresh an element(image) in jQuery
Your click handler is on any img this means that the #currentimage element is the only element which this handler is added. You are then setting the src to the src that it already is.
you instead want to change the src to something else. Here I set an initial smaller placeholder which is replaced with a larger one on click.
$("img").click(function() {
this.src = 'https://placeimg.com/500/500';
$("#imagedesc").html(this.alt);
});
<script src="https://code.jquery.com/jquery-3.4.1.slim.min.js"></script>
<img id="currentimage" src="https://placeimg.com/200/200" alt="ocean vacation" />
<p id="imagedesc"></p>
I am trying to set the image source of an image object with javascript.
I have tried
<img src="" id="image" alt="">
<p id="change">Change</p>
$("#change").click(function () {
$("#image").attr("src", "some source");
});
The problem is that it seems it does change the src, but it doesn't work if the image wasn't loaded when the page loaded. So I cannot suddenly change to the logo of Stackoverflow. Can this be true? How can I then load the image while changing the source?
I had an issue similar to this a couple weeks ago, the easiest way I managed to solve the issue was by loading all image's you will need, and set all of the ones you don't need right away as hidden. Then when your action that causes the change is triggered, you can change the image visibility on the one you want to disappear and the one you want to be shown.
If you have bootstrap installed, it is as easy as adding and removing a class via jQuery
HTML
<img src="" id="image1" alt="">
<img src="" id="image2" alt="" class="hidden">
<p id="change">Change</p>
Javascript
$("#change").click(function () {
$('#image1').addClass('hidden');
$('#image2').removeClass('hidden');
});
At this current iteration, I have it where if a user clicks on an image thumbnail, the thumbnail image displays in a different div (the main div) and in doing so, it rewrites the main div img src attr with the thumbnail img attr minus the "-thumbnail". This part is good, or at least I believe it is.
With that said, after the user clicks on a thumbnail and the main div img appears, it lingers...and what I mean by lingers is that for example, if a user closes the div and re-opens it or another div just like it, the last image shows (stays) when it shouldn't in the main div. Instead, it should be showing the first thumbnail img in the main div...
Any suggestions is appreciated and below is what I currently have (or at least a portion of it). There is a lot more, but below is the main stuff that's giving me troubles...
The HTML part is within a div class called "t_1". I have 24 of these..."t_1", "t_2", "t_3" respectively. And within this class, I have what is posted below in all using the same div classes. The only difference is the folder names in the img tag.
So, when a user clicks on that thumbnail and that thumbnail image gets rewritten so that it can be displayed in the main div "t_main_screenshot", all is good...but then, if the user clicks out of the "t_1" etc. divs, and opens up another "t_2", that last image thumbnail that was clicked previously shows in the main div (t_main_screenshot) instead of the first image thumbnail for what should be in "t_2"...
Hopefully this is a bit better in clarity...It's kind of hard to explain.
HTML:
<div class="t_main_screenshot">
<img src="_framework/images/slides/simplicity/2.png" alt="" title="" />
</div>
<div class="t_thumbnail_wrapper">
<div class="t_thumbnail active">
<img src="_framework/images/slides/simplicity/2-thumbnail.png" alt="" title="" />
</div>
<div class="t_thumbnail">
<img src="_framework/images/slides/simplicity/4-thumbnail.png" alt="" title="" />
</div>
<div class="t_thumbnail">
<img src="_framework/images/slides/simplicity/6-thumbnail.png" alt="" title="" />
</div>
</div>
JS/JQuery:
$('.t_thumbnail').click(function() {
$('.t_thumbnail').removeClass('active');
$(this).addClass('active');
var thumbNail = $(this).find('img').attr('src');
$('.t_main_screenshot img').fadeOut(0, function() {
$(this).fadeIn().css('animation','scale-in .75s ease-in 0s forwards')[0].src = thumbNail.replace('-thumbnail', '');
});
});
Try this code :
$('.t_thumbnail').click(function() {
$('.t_thumbnail').removeClass('active');
$(this).addClass('active');
var thumbNail = $(this).find('img').attr('src');
$(this).parent().parent().siblings().find('.t_main_screenshot').children('img').fadeOut(0, function() {
$(this).fadeIn().css('animation','scale-in .75s ease-in 0s forwards')[0].src = thumbNail.replace('-thumbnail', '');
});
});
As you are updating directly on .t_main_screenshot class. So it will update all places, in this case in all your modals .t_1, .t_2, ...etc.
$('a.screenshots_1').click(function() {
$('.t_main_screenshot img').find('img').attr('src' , "");
});
This will help you to initialize the main screenshot on modal screenshot icon click.
I have an image in my webpage i want when the user click on the image the current image also appear in the other div i have in my web page i am totally confuse please help me out here is my code.
<div id="img"><img src="download1.jpg"></div>
<div><img src="" class="img2"></div>
<img src="download1.jpg" alt="" class="imgs" id="the_id" width="180" height="60" />
<script type="text/javascript">
$(document).ready(function() {
$('.imgs').click(function(){
var idimg = $(this).attr('id');
var srcimg = $(this).attr('src');
alert('ID is: '+ idimg+ '\n SRC: '+ srcimg);
$(".img2").attr('src',srcimg);
});
});
</script>
Fiddle
Tried using a fiddle and it works, here is the code which I used.
HTML
<div id="img"><img src="https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcTgAVjAiDPV-oy-o9mvdHEsVoACUKJIuFcn08RJ5kRYINY4oqjPcA"></div>
<div><img src="" class="img2"></div>
<div>click me </div>
<img src="https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcTgAVjAiDPV-oy-o9mvdHEsVoACUKJIuFcn08RJ5kRYINY4oqjPcA" alt="" class="imgs" id="the_id" width="180" height="60" />
Jquery
$(document).ready(function() {
$('.imgs').click(function(){
var idimg = $(this).attr('id');
var srcimg = $(this).attr('src');
alert('ID is: '+ idimg+ '\n SRC: '+ srcimg);
$(".img2").attr('src',srcimg);
});
});
your code is perfect and will work check out you jquery library that is loaded correctly or not I have not seen any error in your code.
You can achieve this by setting the destination image's src to the same as the img src in the clicked div. Try:
$('.source-img').on('click', function(){
var src = $(this).children('img').attr('src');
$('#destination').children('img').attr('src', src);
});
Check out this jsFiddle.
If you want the images to not have wrappers, then just change the jQuery targetting and leave out the .children(img) part.
If you're looking for a better method, see this JSFiddle
It uses the clone function
$('.imgs').click(function(){
$(this).clone().appendTo('#LoadImageHere').attr('id', 'ClonedImage');
});
It means you don't have an image with an empty src attribute lying around, and you can add attributes to the cloned image as you wish (I added an ID in the example).
If you try width this code or here combination you can get very interesting result.Plz look" . Div element have in self img tag width source image,class "img2". Other image in gallery must have equal id, like is id "the_id". And if we clicking any picture that picture is show in space of div tag. How you can get bouncing effect picture must have equal size. I make combination two div tag where firs have class "img2", and second where is my picture distribute my pictures. With this first code on this page I make nice image gallery. Try and you self, this is not hard.
I am trying to load a video onclick a image in my web page. For that i used the following code.
<script type="text/javascript">
(function($) {
$('a.newID').click(function(){
$('#newID').html('<iframe src="http://www.youtube.com/embed/TJ2X4dFhAC0?autoplay=1" frameborder="0" class="slide" allowtransparency="true" style="width:512px; height:288px;" id="ifm" title=""></iframe>');
});
}(jQuery));
</script>
The html code:
<a href="#" title="" id="newID"><span class="play_icon"><img src="img/play_overON.png" alt=""></span>
<img src="images/slider1.jpg" alt="" class="slide" />
</a>
If i click the image the image will be replaced by the iframe video. What i need is to display a loading icon until the video is loading. How to do that?
Please add following css properties in your click function:
#newID {background-image:url(http://mysite/myloadingimage.gif)}
You will also need to add css property display:inline-block; to #newID, if the display is inline (default).
Replace http://mysite/myloadingimage.gif with actual loading image. Use animated gif for the nice loading effect.
Loading image needs to be centered using css background property. This may depend on size of image.