I have a div that after I click an element(button1), expands its height. I then have another button appear which allows you to shrink the div(button2). After I click button1, the div expands and button2 shows at the bottom. I can then click button2 to shrink the div back to normal, but if I expand the div again, button2 is now off the edge of the screen, albeit in the same bottom location, just far left rather than centered.
I had to set the margin on button2 to -25px since the absolute positioning was kicking it off-center. And I need to use absolute positioning since it seemed it was only way to get the button to appear at the bottom of div after it had expanded.
$(".button1").on("click", function(){
$(".button1" ).fadeOut(200);
$("#block3").animate({
height: '800px'
}, 600, function() {
$(".button2").fadeIn(200);
});
});
$(".button2").on("click", function(){
$(".button2").fadeOut(200);
$("#block3").animate({
height: '400px'
}, 600,function(){
$(".button1" ).fadeIn(200);
});
});
.button2{
width: 50px;
height: 50px;
display:none;
position: absolute;
margin-left: 25px;
}
.button1{
width: 50px;
height: 50px;
margin-top: 25px;
image-rendering: auto;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="row">
<div class = "col-md-12" id = "block3">
<img src="https://placehold.it/120x80/00aaaa/fff/?text=scroll.png" class = "button1" />
<img src="https://placehold.it/120x80/00aaaa/fff/?text=scroll1.png" class = "button2" />
</div>
</div>
Uploaded the code.
https://jsfiddle.net/bs9xhe5e/2/
On the second click it changed the inline styling to display:block instead of display:inline, if you just add the display:inline into your jQuery it works;
https://jsfiddle.net/havL1z3m/
Added
$(".button2").css({display:"inline"});
The reason .button2 is showing up all the way to the left is that the absolute positioning gives it a default left of 0px.
Probably the easiest fix would be to remove absolute position, as well as the margin-left. Instead, to get .button2 to the bottom of the section, just set margin-top: 725px.
.button2{
width: 50px;
height: 50px;
display:none;
margin-top: 725px;
}
And then you will no longer need to use .css() to change bottom of .button2.
Check out this working fiddle: https://jsfiddle.net/gkpx7L15/
Change button2 styles like this:
.button2{
width: 50px;
height: 50px;
display:none;
position: absolute;
margin-left: -25px;
left:50%;//added style
}
https://jsfiddle.net/bs9xhe5e/3/
And everything will work fine
NOTE: Element will be positioned to the center of parent like this only when width is fixed like here. Example:
.someItem{
width: someWidth;//here we set some width to element
left: 50%; //Here we set position from the left
margin-left: -(somewidth/2);//here we set -half of the element's width for margin-left
}
Related
What I am trying to achieve is a tooltip like positioning of an element that's inside the same contain as the element that when clicked will display a div that contains a table.
The full code can be found here:
http://jsbin.com/xihebol
When somebody clicks on .child-table-link the .child-data-table-container is supposed to be positioned relative to the clicked link.
I am doing this inside the click handler of the .child-table-link:
var _this = $(this);
$(".child-data-table-container").css({
"left": _this.offset().left - ($(this).width / 2),
"top": _this.offset().top
})
top works but left doesn't. How I do make it work? In general way to position an element relative to another element in the view-port is what I am looking for.
The problem is you are calculating the left position wrong, "left" will be the position of the left edge of the tooltip container, so if you want it to sit just to the left of your link, it only needs to be set to the negative width of the tooltip container. Check this out:
.canvas {
display: flex;
align-items: center;
justify-content: center;
}
.linkContainer {
position: relative;
background: #000;
}
.tooltip {
position: absolute;
width: 100px;
height: 100px;
background: green;
top: 0;
left: -100px;
}
<div class='canvas'>
<div class='linkContainer'>
<a href='#'>LINK</a>
<div class='tooltip'></div>
</div>
</div>
Notice how the tooltip width is 100px, and the left is set to -100px.
I want a slide effect on a div from left to right or from right to left as in
$('#div').show('slide', {direction:'left'}, 1000);
being my html is
<div id="div-pre">
</div>
<div id="div">
</div>
<div id="div-nex">
</div
But the problem with this approach is that we are hiding the #div initially by setting
#div{
display:none;
}
so that we cannot preserve the width of #div
I have came across another method by making the visibility: hidden as in
$("div").css("visibility", "hidden");
to preserve the width of the div
but this method does not give the sliding effect from left to right or right to left
So I want to achieve both "the effect as in .show('slide', [option], [speed]) altogether with
preserving the div width"
Having no example code to go off, I decided to write a basic example of how you could approach this. Basically, you put an overflow: hidden container around the thing that you want to slide to the left while preserving width, and you then animate a movement leftwards using animate('left':'-pixels');. Your div has to be positioned relatively for this to work. See example below.
$(document).ready(function(){
$('.slideLeft').click(function(){
$('.slider').animate(
{'left':'-600px'},
1000,
function(){
$('.slider').hide();
}
);
});
});
.slider{
height: 300px;
width: 600px;
font-size: 20px;
background-color: yellow;
position: relative;
}
.container{
border: 1px solid red;
height: 300px;
width: 600px;
background-color: silver;
overflow: hidden;
}
.slideLeft{
margin-top: 10px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="container">
<div class="slider">
Hi, I have some content!
</div>
</div>
<button class="slideLeft">Slide me left!</button>
Good luck!
You can wrap your div in another div with overflow:hidden and than you move to right or left the div inside.
I am making a simple reactjs app where I need to put a button over image.
My html looks like:
<div className={"panel-body"}>
<img className={"img-responsive center-block"}
src={album.photos.data[0].source} />
{(this.state.isMouseInsideID === album.id) ? <button>Your Button</button> : null}
</div>
Its all fine except button is shown below the image.
But I want to show the button over the image or in the middle of the div
How can I make this work ?
make button position relative and use z-index: maybe it will be helpful for you.
If you want to center the button in a div, while there is a background image to the div. I would suggest, to assign a background image for the div, rather than inserting an image into the div. Check out the fiddle:
http://jsfiddle.net/49s505sa/1/
HTML:
<div id="wrapper">
<button type="button">Your Button</button>
</div>
CSS:
#wrapper {
width: 100%;
height: 400px;
border: 1px solid black;
background: url('http://placehold.it/350x150') /*assign image, for demo purposes */
}
button {
height: 20px;
position: relative;
margin: -20px -50px;
width: 100px;
top: 50%;
left: 50%;
}
So, inside the render method
var style = {
backgroundImage: 'url(' + album.photos.data[0].source+ ')'
}
<div className={"panel-body"} style={style}>
{(this.state.isMouseInsideID === album.id) ? <button>Your Button</button> : null}
</div>
In this way, we will dynamically assign images to particular divs. And wont have to worry too much about styling.
Hope that helps!
Use z-index properties
The z-index property in CSS controls the vertical stacking order of
elements that overlap. As in, which one appears as if it is physically
closer to you. z-index only effects elements that have a position
value other than static (the default).. Note: z-index only works on
positioned elements
(position:absolute, position:relative, or position:fixed).
img {
position: absolute;
left: 0px;
top: 0px;
z-index: 10;
}
I have two divs with animation, both are doing the same but with different animation. I have
$(document).ready( function(){
$('#show_hide_button').click( function() {
$('#some_box').animate({ width: 'toggle' });
});
});
Whole code in in jsfiddle http://jsfiddle.net/xLHb8/192/
Can anyone please explain to me why first div is animating right to left, left to right and second div is animating always to top left corner.
How can I make second div animate same as first div?
First, the relevant details in your code should be included in your question (in addition to providing the fiddle). But so you have the following CSS:
#some_box {
background: #fc0;
width: 25%;
height: 200px;
}
.second img {
max-width:100%;
max-height:100%;
}
.second {
width: 200px;
}
With the following HTML:
<button id="show_hide_button">click me</button>
<div id="some_box"></div>
<div class="second">
<img src="http://piq.codeus.net/static/media/userpics/piq_66223.png" />;
</div>
Note that you're setting the img to have a maximum width and height of its parent container. So because you're toggling the width of the parent, as parent collapses, the image is scaling down. Further, since you don't have a height setting on the img, its height is going to animate along with the animated width. This creates the effect of the image animating to the top left corner.
Without further details, it's hard to say how to fix your code to achieve the desired effect.
Update
If you want the width only to collapse, you can set a pixel height on your image so that it doesn't scale in proportion to its width:
.second img {
max-width: 100%;
height: 200px;
}
You can also put both animations in a single click event handler, like so:
$(document).ready( function(){
$('#show_hide_button').click( function() {
$('#some_box').animate({ width: 'toggle'});
$('.second').animate({ width: 'toggle' });
});
});
Forked your fiddle: http://jsfiddle.net/u1sdd8j5/1/
Update 2
From the comments, it seems like you want the image to collapse to the left, without losing the aspect ratio. We need to get a little creative to pull that off, especially if you're looking for a solution involving jQuery.animate(). The image actually needs to move downwards as it is scaled down. We can pull that off by animating the <img> itself, rather than its container, and adjusting its top margin at the same time animate its width.
Revised CSS (making the containers the same size for consistency):
#some_box {
background: #fc0;
width: 25%;
height: 200px;
}
.second {
width: 25%;
height: 200px;
}
.second img {
max-width: 100%;
max-height: 100%;
}
Revised JS:
$(document).ready( function(){
$('#show_hide_button').click( function() {
$('#some_box').animate({ width: 'toggle' });
var $secondImg = $('.second img'),
secondImgMargin = $secondImg.is(':visible') ? '50%' : 0;
$('.second img').animate({
width: 'toggle',
marginTop: secondImgMargin
});
});
});
Note that we need to first determine whether or not the <img> is visible. If it is, then we want to animate the top margin to 50%. If it's not, then switch the top margin back to 0.
Here's a new forked fiddle: http://jsfiddle.net/xwanm9ze/1/
Final Note
All of this might be easier to achieve with CSS3 transitions. You would want to set up a class that toggles the animation. And you can specify the transform-origin which, in this case, would be 'left center'.
The problem is, that you added a relative width and height attribute to the inside the second div and did not give a height and width attribute to the second div. This way, the image controls the height and width of the second div, since it has no height and width attribute.
In your case, a solution would be to give the second div a fixed width and height
Also, for the JQuery, you only need one $(document).ready function
$(document).ready(function () {
$('#show_hide_button').click(function () {
$('#some_box').animate({
width: 'toggle'
});
$('.second').animate({
width: 'toggle'
});
});
});
#some_box {
background: #fc0;
width: 25%;
height: 200px;
}
.second img {
width:100%;
height:100%;
}
.second {
width:200px;
height: 200px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button id="show_hide_button">click me</button>
<div id="some_box"></div>
<div class="second">
<img src="http://piq.codeus.net/static/media/userpics/piq_66223.png" />
</div>
I have html sturcture
<div id="bg" class="layer">
<img id="trackmap" src="images/back_2416.jpg" width="1208" height="768" class=" ui-draggable map-icon" usemap="#main-map" data-zoom-image="images/background_zoom.jpg" data-big="images/background_zoom.jpg" style="position: relative; left: -439px; top: -272.6px; margin: 0px; display: inline-block; height: 1327.2px; width: 2088px;">
<div id="nav-text">LOREM IPSUM.</div>
</div>
Jquery
var windowHeight = $("#trackmap").height();
var windowWidth = $("#trackmap").width();
var text_height=((windowHeight)-(100));
$("#nav-text").css("top",windowHeight);
Css
.layer {
position: absolute;
width: 1208px;
height: 768px;
}
#nav-text{
z-index: 200;
color: white;
position: absolute;
font-size: 10px;
margin-left: 715px;
width: 310px;
height: 10px;
position: fixed;
bottom: 5px;}
I just want to fix the nav-text to the bottom right whatsoever.. Now i problem i am facing is theres zoom function on the trackmap.. which increases the height and width of the image ..so the text comes in between of the image ..intereferring with the image.. I have tried taking the image width height using jquery ..but somehow its not working
I am not sure I am following your issue here, but it sounds like you are trying to get a div to be in the bottom-right of another div no matter what size it is. That can be done by setting the parent div position to relative which you have, and the child div position to absolute. You have that set but then override it by setting the position to fixed lower in the CSS. You will also want to set the bottom to 0 and the right to 0.
This will position the child div to the bottom right of the parent div. Then you can get rid of your jQuery. Hopefully this helps.
Ok.. I am in a hurry to catch the bus.. but here's a fiddle that illustrates the idea..
basically you will need to use the scrolltop and left parameters to do so:
$(".container").on("scroll", function() {
$(".nav-text").css("top", $(this).prop("scrollTop") + 130);
$(".nav-text").css("left", $(this).prop("scrollLeft") + 120);
});
but move the scrolls first.. sorry I need to go now..
You can achieve this by not fixing the .layer width and height, using display:inline-block; to prevent the div from filling the whole container width. At that point, the .layer size will match the image size whatever it is.
Finally you just need to set the text to absolute position and bottom and right properties too.
.parent{
display:inline-block;
position:relative;
}
.children{
position:absolute;
bottom:0;
right:0;
}
Here is the fiddle explaining
And here is the proof it works even if the image size is changed(click on the image).
Fiddle 2