Zara's product image functionality - javascript

I'm trying to replicate Zara's product functionality: (click on product image)
https://www.zara.com/es/en/woman/outerwear/view-all/tweed-coat-with-metal-buttons-c733882p5205048.html
I have it almost working from a jsfiddle I found:
https://jsfiddle.net/y6p9pLpb/1/
$(function() {
$('.product-wrapper a').click(function() {
$('.image-zoom img').attr('src', $(this).find('img').attr('src'));
$('.image-zoom').show();
$('.product-wrapper').css('display', 'none');
return false;
});
$('.image-zoom').mousemove(function(e) {
var h = $(this).find('img').height();
var vptHeight = $(document).height();
var y = -((h - vptHeight) / vptHeight) * e.pageY;
$('div img').css('top', y + "px");
});
$('.image-zoom').click(function() {
$('.image-zoom').hide();
$('.product-wrapper').css('display', 'block');
});
});
There's only one thing I can't solve.
When I click on the image and expands, this one jumps to match its respective position relative to the cursor.
Is there any way I can solve that? Like Zara does...
Thanks in advance!

The issue is that the image jumps to your cursor once you move it, right? This should do what you're looking for!
https://jsfiddle.net/8z67g96b/
I just broke out the position change as a function, and also call it when you click the thumbnail - that way, the image already has the cursor's position influencing it when it first appears.
function zoomTracking(event){
var h = $('.image-zoom img').height();
var vptHeight = $(document).height();
var y = -((h - vptHeight) / vptHeight) * event.pageY;
$('.image-zoom img').css('top', y + "px");
}
$('.product-wrapper a').click(function(e) {
$('.image-zoom img').attr('src', $(this).find('img').attr('src'));
$('.image-zoom').show()
zoomTracking(e);
$('.product-wrapper').css('display', 'none');
return false;
});
$('.image-zoom').mousemove(function(e) {
zoomTracking(e);
});

In the fiddle, it has a section relating to the position of the mouse and the width/height of the image:
$('.image-zoom').mousemove(function(e) {
var h = $(this).find('img').height();
var vptHeight = $(document).height();
var y = -((h - vptHeight) / vptHeight) * e.pageY;
$('div img').css('top', y + "px");
});
Simply remove this portion and it won't do any moving with the mouse, just expand onClick.

Related

How to position the center of a div to the center of the mouse cursor on mouse movement with JS?

I'm trying to position the center of a div element to the center of the mouse cursor, that will follow along its movements.
Already I came up with the code below, but the problem with this one is, that the following div is not positioned at the center of my cursor, but with some offset off the cursor.
WORKFLOW
The basic idea behind my code is, when the mouse enters the .post-entry div element, the .pointer within the current item should be displayed and follow the cursor of the mouse. When the mouse leaves the div it should be hidden.
CODE
HTML post item:
<article class="col-md-4 col-sm-6 post-entry">
<a href="#" title="">
<figure class="post-thumb">
<img src="http://placehold.it/300x300" alt="">
<div class="pointer" style="background: red;"></div>
</figure><!-- End figure.post-thumb -->
</a>
</article><!-- End article.col-md-4 post-entry -->
JS:
$('.entry .post-entry').each(function() {
$(this).on("mouseenter", mouseEnter);
$(this).on("mousemove", mouseMove);
$(this).on("mouseleave", mouseLeave);
});
function mouseEnter(event) {
console.log('enter');
var target = $(this);
var dot = target.find('.pointer');
var mX = (event.clientX);
var mY = (event.clientY);
set(
dot, {
x: mX,
y: mY,
force3D: !0
}
);
};
function mouseMove(event) {
console.log('move');
var target = $(this);
var dot = target.find('.pointer');
// var offset = target.offset();
// var width = target.width();
// var height = target.height();
// var top = offset.top;
// var left = offset.left;
var mX = (event.clientX);
var mY = (event.clientY);
$(dot).css('-webkit-transform', 'translate3d(' + mX + 'px, ' + mY + 'px, 0)');
};
function mouseLeave(event) {
console.log('leave');
var target = $(this);
var dot = target.find('.pointer');
$(dot).css('-webkit-transform', 'translate3d(0, 0, 0) scale(0, 0)');
};
function onClick(event) {
event.preventDefault();
console.log('click');
};
function set(el, obj) {
var dot = $(el).css('-webkit-transform', 'translate3d(' + obj.x + 'px, ' + obj.y + 'px, 0px)');
return dot;
};
PROBLEM / DEMO
As mentioned before, the span is following the mouse cursor, only the span is not positioned to the center of the cursor. It will be offset the mouse. See live demo here
I tried already something like this for the mX and mY variables, but with no succes:
var mX = (event.clientX - $(this).offset().left) / $(this).width() * $(this).width() - .125 * $(this).width();
var mY = (event.clientY - $(this).offsetTop) / $(this).height() * $(this).height() - .125 * $(this).width();
Also the answer from #hiEven doesn't work and will let me with the same issue:
transform: calc(mX - 50%, mY - 50%)
I know I should do something with dividing the .pointer by half, but how I should implement that in the code is a big question mark for me.
UPDATE
I created two new Codepen projects:
Use without images: http://codepen.io/anon/pen/GqGOLv. When you hover over the first item you will see that the brown pointer is correctly following your mouse cursor - what I am looking for. But when hovering over the second one, you will see the red pointer, only when you are at the very left side of the item.
When I use images: http://codepen.io/anon/pen/QExOkx. The problem by this example is that when you at the very top of the first column, you will see the brown pointer. When hover at the top left corner of the second item you will see a little piece of the red pointer, the same as the example without images.
Both pointer should follow the mouse cursor correctly. And I am searching for a solution that works with the use of an image.
Beside these two examples, when I add to the first one, an extra margin-left to the first item, the brown pointer will not be in the center of the mouse cursor, only when it's set to margin-left zero.
So I don't know what's missing and why it only works with the first example (without images) and only for the first item?
Try the code below
<html>
<head>
<style>
#mouse_div{
position: absolute;
background-color: black;
}
</style>
<script>
var div_width = 100;
var div_height = 100;
var div_x, div_y;
function mouse_position(event){
var mouse_x = event.clientX;
var mouse_y = event.clientY;
document.getElementById("mouse_div").style.width = div_width + "px";
document.getElementById("mouse_div").style.height = div_height + "px";
div_x = mouse_x - (div_width / 2);
div_y = mouse_y - (div_height / 2);
document.getElementById("mouse_div").style.left = div_x + "px";
document.getElementById("mouse_div").style.top = div_y + "px";
}
</script>
</head>
<body onmousemove="mouse_position(event)" onload="mouse_position(event)">
<div id="mouse_div"></div>
</body>
</html>
This program gets the position of your mouse, the width, and the height of the div. Then, it takes the x and subtracts the div's width divided by two from it (this centres the div's x position on your mouse). The program then does the same thing for the mouse y. Once all of the variables are defined, I use JavaScript to access the CSS of the div to place the div where it needs to be.
Note: you must make sure that the position of the div is set to absolute or the program will not work.
I assume you want the circle being center of your mouse, right?
try do this
transform: `translate(calc(${mx}px - 50%), calc(${my}px - 50%))
here is the demo
Based on my latest update, I did not conform to the correct formula that is needed to center the element .pointer to the mouse.
In order to use the following calculation within mouseMove:
var mX = (event.clientX);
var mY = (event.clientY);
Should be changed to this:
var height = dot.height();
var width = dot.width();
var offset = target.offset();
var w = target.width();
var h = target.height();
var top = offset.top;
var left = offset.left;
var mX = (event.clientX - left) - width / 2 - 15; // 15 = padding
var mY = (event.clientY - top) - height / 2;
So this formule is considering that the following DOM element .pointer will follow the mouse movements of the user. I don't know exactly why this working, but the offset from the previous item will be decreased from the current clientX coordinates, so the position of the second item is reset to zero, so the pointer will start at the left side of each item.
Here is a working demo of above code: http://codepen.io/anon/pen/AXdxZO?editors=0110

Javascript moving child when dragging on parent

I want to move child element when dragging on parent and the element itself, and the parent shouldn't be moved.
here is my demo
As in the demo, I want to move the red box when either drag on it or drag on its parent (the background), but I couldn't compute the right position, could you help me?
Other question is why I can't offthe mousemove event when I set .off('mousemove', mousemove')
Thank you very much
I have done some changes. This works ;)
DEMO
$(function(){
var graph = $('.graph')[0];
var parent = $(graph).parent();
var lockX = 0;
var lockY = 0;
var mousemove = function(e){
$(graph).offset({
top: e.pageY + lockY,
left: e.pageX + lockX
});
};
parent.on('mousedown', function(e) {
lockY = $(graph).offset().top - e.pageY;
lockX = $(graph).offset().left - e.pageX;
$(this).addClass('draggable')
.on('mousemove', mousemove)
.on('mouseup', function(){
$(this).off('mousemove', mousemove)
})
event.preventDefault()
});
$('.graph').parent().on('mouseup', function(e) {
$('.draggable')
.off('mousemove', mousemove)
.removeClass('draggable');
});
});
EDIT
The only actual change I made to make it work was:
.offset({
// instead of: e.pageY - $('.draggable').outerHeight() / 2 + dtop
top: e.pageY + dtop,
// instead of: e.pageX - $('.draggable').outerWidth() / 2 + dleft
left: e.pageX + dleft
})
PS: In the example in the JS Fiddle I changed dleft and dtop respectively to lockX and lockY. Ofcourse, that is a pure semantical thing.

Finding Image Height variable for multiple images in Jquery

I'm currently working on this script for "tooltips" on a website. I'm finding that the code I currently have will get the image height for my first tooltip image on the page ('pop1') but it ignores the rest (they come out as null).
What's the most effective way to get all the tooltip image heights, and use them every time the user scrolls over the tooltip image?
Another issue, if anyone is able to figure this one out - is that on my FULL webpage (many more divs, rows, columns, etc.) the script begins to break because clientX and clientY are being affected by the various divs and page elements.
I'd like to be able to set clientX and clientY to the exact (x, y) coordinates that the user's mouse is at, relative to the entire webpage, not relative to the page's child elements.
Thanks
Here's my JSFiddle: http://jsfiddle.net/tgs7px4f/18/
JS Code:
$('a.popper').hover(function (e) {
var target = '#' + ($(this).attr('data-popbox'));
$(target).show();
}, function () {
var target = '#' + ($(this).attr('data-popbox'));
if (!($("a.popper").hasClass("show"))) {
$(target).hide();
}
});
$('a.popper').mousemove(function (e) {
var target = '#' + ($(this).attr('data-popbox'));
// images vary in height!
// images are all 366px wide.
var imageWidth = 366;
var imageHeight = $(".popimg").height();
//alert('Image Height: ' + imageHeight);
//Offset tooltip:
//10px to the right of cursor
var imageX = e.clientX + 20;
//imageHeight up from cursor
var imageY = e.clientY - imageHeight - 20;
// Find bounds of current window, and if...
// Tooltip goes off right side:
if ((imageX + imageWidth) > $(window).width()) {
//Move tooltip left so it meets edge:
imageX = $(window).width() - imageWidth;
}
// Tooltip goes off top
if (imageY < 0) {
//Move tooltip down so it meets top:
imageY = 0;
}
$(target).css('top', imageY).css('left', imageX);
});
What's the most effective way to get all the tooltip image heights, and use them every time the user scrolls over the tooltip image?
First of all, I suppose you mean whenever a user does a mouseover on one of the elements? However, this seems to work and it cashes the height of the image directly on the element and uses it the next time a mouseover occurs:
$('a.popper').mousemove(function (e) {
var target = '#' + ($(this).attr('data-popbox'));
// images vary in height!
// images are all 366px wide.
var imageWidth = 366;
$target = $(target);
if (!$target.attr("height")) {
var img = $target.closest(".popbox").children("img");
var imageHeight = img.height();
$target.attr("height", imageHeight);
console.log("height attribute set");
} else {
var imageHeight = +($target.attr("height")) + 0;
console.log("cached height used");
}
console.log('Image Height: ', imageHeight);
//Offset tooltip:
//10px to the right of cursor
var imageX = e.clientX + 20;
//imageHeight up from cursor
var imageY = e.clientY - imageHeight - 20;
// Find bounds of current window, and if...
// Tooltip goes off right side:
if ((imageX + imageWidth) > $(window).width()) {
//Move tooltip left so it meets edge:
imageX = $(window).width() - imageWidth;
}
// Tooltip goes off top
if (imageY < 0) {
//Move tooltip down so it meets top:
imageY = 0;
}
$(target).css('top', imageY).css('left', imageX);
});
Obviously, you should remove all the console.log statements which are for testing purposes only.
jsFiddle
Regarding your second question, it's hard to say anything concrete without another jsFiddle or additional code.

jQuery Change Background Position Using Relative Value

I am trying to change the image background position X & Y using relative values (+= and -=) when click on the button. It seems to reset itself to 0% 0%. It didnt work.
Fiddle
$(function(){
$("body").on("click", "#move", function() {
$("#obj-1").css({ 'backgroundPosition': '+=0px -=5px' });
return false;
});
})
Change the click function as below:
$(function(){
$("body").on("click","#move", function(){
var backgroundPos = $("#obj-1").css('backgroundPosition').split(" ");
var xPos = parseInt(backgroundPos[0], 10);
var yPos = parseInt(backgroundPos[1], 10);
var newX = xPos + 0;
var newY = yPos - 5;
$('#obj-1').css({
'background-position':newX+'px '+newY+'px'
});
return false;
});
});
Updated fiddle here. hope it helps.
Now on click image will shift from top and left to 10px eveyrtime.
here is the updated jquery code.
$(function(){
$("body").on("click","#move", function(){
$("#obj-1").css({
left: $("#obj-1").position().left + 10 + "px",
top: $("#obj-1").position().top + 10 + "px"
});
return false;
});
})
Here is the working Demo http://jsfiddle.net/kheema/8f2pf/7/

Basic Parallax - Background jerking up

I have a simple code for moving the background images, but the image is jerking up once the section is in view. the idea is for the background-pos to move only when the section is in view.
Any ideas?
$window = $(window);
$('.portfolioSection').each(function(){
var $bgobj = $(this); // assigning the object
var speed = 8;
$(window).scroll(function() {
if($(window).scrollTop() + 150 >= $bgobj.offset().top){
// Scroll the background at var speed
// the yPos is a negative value because we're scrolling it UP!
var yPos = -($window.scrollTop() / speed);
// Put together our final background position
var coords = '0 '+ yPos + 'px'
// Move the background
$bgobj.css({ backgroundPosition: coords });
}
}); // window scroll Ends
});
Solved it- I didn't account for the section position - Line 9
var yPos = -(($window.scrollTop()-$bgobj.offset().top) / speed);

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