How can i get the position of an element relative to its parent?
Let's suppose we have an image object called 'clone', and a div as its parent:
var parent = clone.parent().offset();
var pos = clone.offset();
If i drag the image right in the top left corner of the div i will get as result almost the same values. But what about if i do this:
var top = pos.top - parent.top;
var left = pos.left - parent.left;
it'll give me the relative position of the image to its parent, right? But what happen if i have to store those values and display them on a different browser size?
The answer it's pretty easy, they won't show inside the div because the position may change.
After this little explaination, my question is:
Is there a way to avoid this kind of problem, and directly take the relative position to its parent?
This is the actual code i'm using, so you can give a look, and try to figure out how to make it works properly! http://dpaste.com/3M1MGQW
Have you set position: relative; to the parent so the positions are relative to it?
Well, any possitioning other tha static will work, but relative without specifing top and left may be the most painless
Related
Is it possible to keep the top row moving like we move the first column in a table using jQuery?
The code I used to keep the first column moving during scroll is something like this.
$('#table-name').scroll(function () {
var _left = $(this).scrollLeft();
$('.firstTd').css('left', _left);
});
when I use the same technique to top property to a table row...through the CSS gets applied it is not honored by the browsers.
P.S: I used left property on td element and want to apply the same technique to a tr tag
Demo here: https://jsfiddle.net/8w4qac30/7/
EDIT
Oops, understood the question bad. I'll keep the info below, but actually my answer is this.
As trs are quite picky, the only thing I can think of is to select all the tds and move them, like you do with the first one, like this: https://jsfiddle.net/8w4qac30/9/
Old answer
left, top, right and bottom are positioning attributes, and for them to work you should set the position attribute too.
position attributes come in different flavors:
relative means to position the element relative to itself, so if you add, for example, left: 20px to a relative positioned element, it simply will shift its position 20 pixels to the left.
absolute means to position the element relative to the first parent that is also relative or absolute positioned.
fixed means to position the element relative to the browser window and will keep fixed during scrolls without additional code. I think that you should go this way.
Check this:
Check the positions here: https://www.w3schools.com/cssref/pr_class_position.asp
I'm trying to create a spot the difference game with jQuery.
Basically, several images stacked, positioned absolutely in a container. Above the container there is the page header with a logo and a menu, which takes altogather about 120px above the images container.
When someone clicks an area inside the image, I put there a new div, with either a correct (V) mark, or a wrong (red X).
I'm trying to get the position of the click inside the element, using the following code (the following used event variable e is returned in the click event just to be clear):
var parentOffset = $(this).parent().offset();
var topOffset = e.clientY - offset.top;
My problem is that the offset from the top changes when I scroll the page down a little to the footer area, and then I do not position the new marker div in the correct height.
When I'm scrolled to the top of the page, the mark is position correctly.
I've created such a game before, but can't understand why suddenly the calculation is wrong :\
Seems that I get the distance minus the scroll height, but not sure.
Thanks for your insight,
Yanipan
I played around with Firebug a little and it looks like e.originalEvent.layerY is exactly what you're looking for.
It always shows the absolute coordinates of your click within the clicked object, no matter where the screen is scrolled.
I have a div with a relative position (div 1). It contains an image (shown as the green block) with an absolute position which by default is hidden way off to the left of the browser window.
I'd like to make it so that when div 1 is in the center of the browser window as the user scrolls down, the image is moved in slightly from the left and appears on the screen. As the user begins to scroll down past div 1, I'd like the image to move back to its original offscreen position.
I have attached a picture to try and make a bit more sense.
I have a feeling this is possible using JavaScript or jQuery but I'm not sure how. Any help would be greatly appreciated.
Ian
You'll want to bind a handler to the scroll event of the window, and measure the ratio of how far down the page the user has scrolled - then, position the image accordingly. I built a rough prototype; you should be able to tweak sizes and positions to make it work for you.
The JS for the prototype, which depends on the HTML and CSS in the JSFiddle linked above, is as follows:
var $main = $('.main');
var $tgt = $('.targetMover');
var origLeft = $tgt.position().left;
var maxLeft = 200;
$main.scroll(function(ev){
var ratio = $main[0].scrollTop / $main[0].scrollHeight;
var newLeft = origLeft + ( (maxLeft - origLeft) * ratio);
$tgt.css({left:newLeft});
});
You would want to position the image on scroll. You would basically check what the position of the div is, set the top of the image to the same as the div, and set the left to whatever you like. you could use jquery animate for this to make it "move" to that position. You then would have to manage to do an scrollstop event (which doesn't exist), and hide the image again. See: http://james.padolsey.com/javascript/special-scroll-events-for-jquery/ for scrollstop implementation (taken from the below post).
You might want to read through jQuery - fadeOut on Scroll / fadeIn on "scrollstop"
We have a script used to edit certain files inline. Basically, each file is broken down into sections, and hovering over a section will bring up a set of tools (just a div with image buttons) that you can use to edit that particular section. We have the parent elements (sections) set as position: relative, and the set of tools set as position:absolute which are set to the right side of the section. This all works fine, especially since many of these are rather small.
However, we do have many of these sections which can become quite large, reaching lengths of two screens or even more. In these cases we would like for the tools to sort of flow with the user's scrolling, so say if the user is looking at the vertical-middle of the section, the buttons will rest at the vertical middle as well, however, if the vertical center of the user's screen scrolls past the section but the user is still hovering over the section, we would like for the tools to remain within their parent element and not be able to pop out.
We already have a script to move an element with the user's scroll if it goes out of the screen, so I was thinking I could modify that a bit to do that, I'm just not sure how to bound the element by it's parent.
TL;DR: How would I create an element that attempts to be vertically centered in the user's window, but cannot leave it's parent element.
Keeping it vertically aligned but only displayed when the section is hovered wouldn't do the trick?
This sounds like a manual positioning. You could use jquery to get the size of the browserwindow, get the scroll offset of the parent and calculate the top of the tools acording to screensize and scrolloffset value of the parent.
I don't thin css can handle this alone.
You could also just use one toolbox for all sections and pass the parent element as parameter.
Best wishes
Here's a quick implementation I came up with based on Andreas's suggestion
$(window).scroll(function(){
var a = $(window).scrollTop() + ($(window).height() * .35);
var b = $("#movedelement");
var c = $(window).scrollTop() + ($(window).height() * .48);
if (a < (b.parent().offset().top + 8))
b.css({position: "absolute", top: "1em" });
else if (c > (b.parent().offset().top + b.parent().height() - 8))
b.css({position: "absolute", top: b.parent().height() - 100 });
else
b.css({position: "fixed", top: "35%" });
}
Tweak some numbers around for the element height. Dirty, but works.
I have a page with many divs and style, with my div buried somewhere inside.
I am trying to have a button that automatically makes my div, that includes a video player, resize and capture the whole browser.
In order to do that I am trying to get the current position of the div and then position it relatively so that it'll get to the top-left corner so I could then use document.body.clientHeight/clientWidth.
Can't get this to work.
I tried the approach of moving my div to the first div and then resizing however this messes up the flash player.
Any ideas? any different approaches?
Thanks,
Guy
Use one of the lightbox clones that can handle DIVs. They usually copy the DIV in question into their own view DIV, which helps with positioning issues and you don't have to do anything to the buried div.
I find Multi-Faceted lightbox to be very easy for customizations:
http://www.gregphoto.net/lightbox/
but there are lots of others around as well.
Why relative?
You should rather use fixed instead of relative. Then set positon to 0,0 and width and height to 100%.
Simple js can do this.
On click, just set the div's style to 'fixed', and position 0,0. Like:
var theDiv = document.getElementById('yourDivsId');
theDiv.style.position = 'fixed';
theDiv.style.top = 0;
theDiv.style.left = 0;
This should do the trick:
<div style="position:absolute;top:0;left:0;width:100%;height:100%">
some content here
</div>