I'm trying to make a div that I have on top of another div show up when you click on something.
This is the code for the two divs, without all the stuff that's within each:
<div id="randomarticle_enlarge">
<h1></h1>
<h4></h4>
<p></p>
<p></p>
</div>
<div class="bodybag">
<h1></h1>
<h4></h4>
<p></p>
<p></p>
</div>
Then I have css for each, of course:
.bodybag {
width:960px;
}
#randomarticle_englarge {
height:750px;
width:960px;
position:absolute;
z-index:2;
margin-top:1px;
padding-left:20px;
padding-right:20px;
display: none;
}
Am I supposed to have the bodybag class have a z-index and a position:relative? Because even though I don't it's working (at this point).
Anyway, I have this script written that's doing exactly what I want it to do:
$(document).ready(function() {
$('.popular').click(function() {
$('#textmask').fadeTo( 'fast', 0.1);
$('#backgroundmask').css('background-color', 'white');
});
});
And all I want to happen next is that as the textmask and the backgroundmask fade in/change as they should and do, is for the randomarticle_enlarge div to show up.
I've tried using .toggle and .toggleClass and .slideToggle and .show but nothing is working.
Absolute positioning must be relative to a container. In order to absolutely position something you need to indicate what it's absolutely positioned to. Something along these lines.
<div id="randomarticle_englargeContainer">
<div id="randomarticle_englarge">
</div>
<div class="bodybag">
</div>
</div>
#randomarticle_englargeContainer {
position: relative;
}
.bodybag {
position: absolute;
left: 0;
top: 0;
}
When copying everything from above I have no issues using $('#randomarticle_englarge').toggle();. Check your browser's console for errors; you might find the answers there.
I'm not exactly sure about what would you like to do with the divs, but I created an example for you, maybe this is what you want:
LIVE DEMO
So there is two divs. The 2nd div covers the 1st one. Clicking on a 'button' hides the 2nd div, so the 1st one reveals. Clicking again the 'button', the 2nd div appears and covers the 1st one again.
HTML:
<div class="popular">Click me!</div>
<div class="container">
<div id="randomarticle_enlarge">
<h1>A</h1>
<h4>B</h4>
<p>C</p>
<p>D</p>
</div>
<div class="bodybag">
<h1>E</h1>
<h4>F</h4>
<p>G</p>
<p>H</p>
</div>
</div>
CSS:
.container {
position: relative;
}
.bodybag {
width: 100px;
height: 200px;
background-color: red;
}
#randomarticle_enlarge {
width: 100px;
height: 200px;
background-color: green;
position: absolute;
top: 0;
left: 0;
}
.hide {
display: none;
}
jQuery:
$(document).ready(function() {
$('.popular').click(function() {
$('#randomarticle_enlarge').toggleClass('hide');
});
});
Related
I have two <div> elements attached to each other, I mean there is no space between them.
<div id="box1">1</div>
<div id="box2">2</div>
And I have this jQuery code:
$('#box1 , #box2').hover(function() {
console.log("Hovered")
}, function() {
console.log("Not")
});
My problem is when I move the mouse between box1 and box2, I still get on console log "Not".
I want those divs to be considered as one element so when I move the mouse between them I don't get on console log "Not".
Thanks in advance!
I want those divs to be considered as one element
Well, quite simply, they aren't. And they can't be. That's not how HTML and CSS works.
The hover event is triggered one for each individual element bound to the event handler. And every time you leave one of those elements it will print the "not" output as per your instructions.
There is no "fix" for this in the exact way you described, but there are alternative approaches. An obvious solution is to wrap them both in an outer div and bind the hover event to that instead. Then the whole area will be considered as one element (because it literally is). Demo:
$('#boxcontainer').hover(function() {
console.log("Hovered")
}, function() {
console.log("Not")
});
#boxcontainer {
border: solid 1px black;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="boxcontainer">
<div id="box1">1</div>
<div id="box2">2</div>
</div>
friend check the code below. I think it will work for you. As you have dai you have an absolute position div you must need a parent div and the parent div position must be relative. For doing that you have to add just a simple CSS code position: relative;. You also need to do some changes to your jquery code. You can just hover on the parent div and it will do your work. Hope this code will help you.
//Box 1 Demo
$('#boxParrent1').hover(function() {
console.log("Hovered")
}, function() {
console.log("Not")
});
//Box 2 Demo
$('#boxParrent2').hover(function() {
console.log("Hovered")
}, function() {
console.log("Not")
});
/*Main Code that are needed*/
#boxParrent1, #boxParrent2 {
position: relative;
}
/*Codes Just used to give you a demo*/
#boxParrent1, #boxParrent2{
display: flex;
margin-bottom: 50px;
border: 1px solid #000;
}
#boxParrent1{
width: 200px;
}
#boxParrent2{
width: 210px;
}
#box1, #box2, #box3, #box4{
background: tomato;
width: 100px;
height: 100px;
display: grid;
justify-content: center;
align-items: center;
font-family: Arial;
font-size: 50px;
color: #fff;
cursor: pointer;
}
#box2, #box4{
position:absolute;
top: 0;
left:100px;
background: #02dce6;
}
#box4{
left:110px;
background: #02dce6;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="boxParrent1">
<div id="box1">1</div>
<div id="box2">2</div>
</div>
<div id="boxParrent2">
<div id="box3">3</div>
<div id="box4">4</div>
</div>
Try to place your 2 div's in one super div
<div id="super">
<div id="box1">1</div>
<div id="box2">2</div>
</div>
$('#super').hover(function() {
console.log("Hovered")
}, function() {
console.log("Not")
});
I'm having a little trouble with CSS and can't seem to find a solution. I have this HTML
<div id="closelink">
Close
Click to close
</div>
Now I want to hide the text «Click to close» only, without hiding neither the div, nor the link within it.
Can this be done?
The visibility attribute can be overriden on children elements, so you are able to do this:
#closelink {
visibility: collapse;
}
#closelink a {
visibility: visible;
}
<div id="closelink">
Close Click to close
</div>
.closelink {
font-size: 0px;
}
.close-link a {
font-size: 12px;
}
Try
#closelink {
position: relative;
left: -9999px;
}
#closelink a {
position: relative;
left: 9999px;
}
<div id="closelink">
Close Click to close
</div>
It works but you can use visibility:hidden instead of visibility:collapse
To avoid the child element breaking to a new line (as happens with just using visibility: hidden/collapse and visibility: visible), and to avoid drawing a 9999px block in the browser (generally frowned upon as it is unnecessary overhead), try this:
#closelink {
position: relative;
visibility: hidden;
}
#closelink a {
position: absolute;
left: 0;
top: 0;
visibility: visible;
}
<div id="closelink">
Close Click to close
</div>
You can adjust your left: 0 value to provide some padding.
There are three methods I could think of:
One
#parent {
opacity: 1;
}
.child {
opacity: 0;
}
<div id="parent">
dkjfdkf
<span class="child">Annoying text</span>
</div>
Two
.child {
visibility: hidden;
}
<div id="parent">
dkjfdkf
<span class="child">Annoying text</span>
</div>
Three
.child {
display: none;
}
<div id="parent">
dkjfdkf
<span class="child">Annoying text</span>
</div>
Opacity is best if you want the image to always be on the page to keep the structure but, you don't want it to be visible.
Hope this was helpful.
I've seen some custom Scrollbard but don't work for what I need...
I have a div element with dynamic text, sometimes there is lots of text so scroll bars shows up, I wonder if is possible to overflow: hidden; and have an image (arrow pointing down) that when clicked, the div will scroll down normally like when using the browsers scrollbar.
I have seen lots of this: https://grsmto.github.io/simplebar, all have scroll bars on the side, none has what I want.
Here it is (only the basics):
function scrollDown() {
var cuttentOffsetTop = $('#inner').offset().top
$('#inner').offset({top: (cuttentOffsetTop - 10)})
}
#container {
width: 100%;
height: 300px;
background-color: gray;
overflow-y: hidden;
position: relative;
}
.item {
width: 100%;
height: 100px;
background-color: violet;
}
.item + .item {
margin-top: 10px;
}
#scroll-down {
background-color: forestgreen;
color: white;
margin-bottom: 10px;
cursor: pointer;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="scroll-down" onclick="scrollDown()">Click here to scroll down</div>
<div id="container">
<div id="inner">
<div class="item">1</div>
<div class="item">2</div>
<div class="item">3</div>
<div class="item">4</div>
<div class="item">5</div>
<div class="item">6</div>
<div class="item">7</div>
<div class="item">8</div>
<div class="item">9</div>
<div class="item">10</div>
</div>
</div>
If you need an explanation - just ask.
Have you actually attempted to create this? Provide code that you have attempted so that we may edit that, as opposed to writing the whole thing for you. You didn't make it entirely clear if you wanted to jump down to a position, slowly scroll down while the button is held down, or what exactly so I'll provide a few different types.
window.scrollTo(0, 100);
If you know how far down you want to jump, you could use this. Alternately, using HTML you can do the following to jump to a specific part of a page.
Jump to element with id jumpLocation
You just have to google it better. Look at element.scrollTop method, more here. And a thread from stackoverflow..
I have 3 fieldsets.
What I would like to make is this layout:
I want the bottom right fieldset to be bottom aligned, so it's bottom would be aligned with the left fieldset.
It should work in different resolutions.
Is there an easy way? or I will have to use javascript to add to it a margin-top dynamically?
code:
<div class="fieldSetsContainer">
<fieldset class="leftFieldSet">test
<br/>test
<br/>test
<br/>test
<br/>test
<br/>test
<br/>
</fieldset>
<div class="rightFieldSets">
<fieldset>test2</fieldset>
<fieldset class="bottomRightFieldSet">test3</fieldset>
</div>
css:
.rightFieldSets {
float:left;
width:34%;
}
.rightFieldSets fieldset {
clear:left;
width:89%;
}
.leftFieldSet {
width:62%;
float:left;
margin-right:1px;
}
.bottomRightFieldSet {
margin-top:6px;
}
here is the a link:
http://jsfiddle.net/bbryK/
My solution assumes two things:
The right column has a fixed width.
The left column must always be the highest.
See http://jsfiddle.net/c3AFP/2/
Html structure:
<div class="container">
<div class="right">
<fieldset class="top"></fieldset>
<fieldset class="bottom"></fieldset>
</div>
<fieldset class="left"></fieldset>
</div>
Css styles:
.container {
position: relative;
}
.top, .bottom {
width: 300px;
}
.left {
margin-right: 300px;
}
.right {
float: right;
margin-left: 10px;
}
.bottom {
position: absolute;
bottom: 0;
}
EDIT:
Here is a solution with the right column sized by percentage: http://jsfiddle.net/c3AFP/5/
EDIT 2:
Here is a table based solution which removes the requirement of the left column being the tallest. Using vertical-align you can adjust where the smaller elements should align in relation to the tallest one: http://jsfiddle.net/c3AFP/7/
I'm giving you a start point on fiddle. Please play around, make some code and do share the same.
http://jsfiddle.net/vY462/
#one{width:200px;height:70px;border:2px solid black;float:left;}
#two,#three{width:200px;height:25px;border:2px solid black;float:right;margin-top:5px;}
<div id="one">1</div>
<div id="two">2</div>
<div id="three">3</div>
What I want to do is to show the top photo (which is set to visibility: hidden) on hover. I have two photos positioned on each other like this:
<div class="frame">
<img src="./img/portfolio/default1.jpg" width="300" height="178" alt="Title Here"></a>
<div class="boxwrapper" style="visibility: hidden;"></div>
</div>
Added the second photo through css:
.boxwrapper {
background: url("../img/boxPlus.gif");
position: relative;
width: 300px;
height: 178px;
left: -6px;
top: -184px;
z-index: 1000;
}
Is it possible to do with css? Tried (and several more options):
#frame img:hover .boxwrapper {
visibility: visible;
}
But it is not working. Or this is only possible with javascript? If yes, please give some tips as I am not too much of javascript guy. Thanks!
You could set the photo as background of the boxwrapper
.boxwrapper{
background: url("../img/boxPlus.gif");
}
.boxwrapper:hover{
background: url("../img/portfolio/default1.jpg");
}
if this is not possible you could add it as background trough a style attribute inside your html
<div class="boxwrapper" style="background: url('../img/boxPlus.gif');" ></div>
You'd have to put the :hover class on a parent container. CSS does not allow such things to trickle "up" the tree, only down.
.boxwrapper {
display: none;
}
.frame:hover .boxwrapper {
display: block;
}