HTML - Display image on hover

Question

So I have this div class:

<div id="object1">
    <div class="title" onmouseover="hoverImage(event)">SPT 3</div>
</div>

And outside of that div, I have another div with a image:

<div class="hover_img" id="hover_img">
    <img src="img/sptgoal.PNG">
</div>

CSS code to center the image:

.hover_img {
    position:relative;
    height: 100%;
    width:100%;
    z-index: 99;
}

.hover_img img {
    position: fixed;
    top: 50%;
    left: 50%;
    transform: translate(-50%, -50%);
    display:none;
}

I only want the image to be visible when hovering the div. This is the Javascript code I have:

function hoverImage() {
    var x = document.getElementById("hover_img");
    x.style.display = "block";
}

But when hovering the mouse over, nothing shows up. What am I doing wrong here?

alexP · Accepted Answer · 2021-02-19 09:32:57Z

2

With document.getElementById("hover_img") you don't show the image, but the container around it (hover_img). But you have hidden the image (.hover_img img) in your CSS.

Use document.querySelectorAll("#hover_img>img") for selecting your image. querySelectorAll will return a NodeList, so you have to use x[0] for selecting the first element in list.

function hoverImage(event) {
  var x = document.querySelectorAll("#hover_img>img");
  var sDisplay = (event.type === "mouseout") ? "none" : "block";
  x[0].style.display = sDisplay;
}

.hover_img {
    position:relative;
    height: 100%;
    width:100%;
    z-index: 99;
}

.hover_img img {
    position: fixed;
    top: 50%;
    left: 50%;
    transform: translate(-50%, -50%);
    display:none;
}

<div id="object1">
    <div class="title" onmouseover="hoverImage(event)" onmouseout="hoverImage(event)">SPT 3</div>
</div>

<div class="hover_img" id="hover_img">
    <img src="https://picsum.photos/200/300">
</div>

edited Feb 19, 2021 at 9:32

answered Feb 19, 2021 at 9:18

alexP

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2 Comments

Carben Over a year ago

Awesome! But why doesn't the image goes away when not hovering anymore?

alexP Over a year ago

There is no event for hiding the image. Check my answer, i've edited it. Now there is a onmouseout event in div tag and in hoverImage function it is checked which event is called

Deepak Singhal · Accepted Answer · 2021-02-19 09:21:27Z

1

problem is on element selector you need to select image element for change display property not div tag.
i hope it's help full

Blockquote

function hoverImage() {
    var x = document.getElementById("hover_img");
    x.children[0].style.display = "block";
}

.hover_img {
    position:relative;
    height: 100%;
    width:100%;
    z-index: 99;
}

.hover_img img {
    position: fixed;
    top: 50%;
    left: 50%;
    transform: translate(-50%, -50%);
    display:none;
}

<div id="object1">
    <div class="title" onmouseover="hoverImage(event)">SPT 3</div>
</div>
<div class="hover_img" id="hover_img">
    <img src="img/sptgoal.PNG">
</div>

answered Feb 19, 2021 at 9:21

Deepak Singhal

441 silver badge4 bronze badges

Comments

parhammed · Accepted Answer · 2021-02-19 09:24:35Z

0

you must select another id for img:

<div class="hover_img" id="hover_img">
    <img src="img/sptgoal.PNG" id="image1">
</div>

and use this id in your java script:

function hoverImage() {
    var x = document.getElementById("image1");
    x.style.display = "block";
}

answered Feb 19, 2021 at 9:24

parhammed

339 bronze badges

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