I have this little problem and I am requesting for your help. I have a div element, inside which I have an img element, like this
<div id="parent" onmouseover="MyFuncHover()" onmouseout="MyFuncOut()">
<img id="child" src="button.png" style="visibility: hidden" />
</div>
<script type="text/javascript">
function MyFuncHover() {
// Here I have some code that makes the "child" visible
}
function MyFuncOut() {
// Here I have some code that makes the "child" invisible again
}
</script>
As you, see, the image is a child of the div. I want that only when I leave the div, the child to disappear. Yet, it looks like when I move the mouse over the image, the MyFuncOut() function is called (because, I suppose, I leave the div by hovering the image). I don't want that to happen. I want the MyFuncOut() function to be called only when I leave the div area.
I didn't know that when you move your mouse over a child control, it's parent calls the mouseout event (even if I am over the child, I am still over the parent, too). I am trapped into this and I need a little of your good advice. Thanks!
CHANGES
O.K. Event bubbling will not send the "mouseout" event to the parent when I "mouseout" the child. It also won't send the "mouseover" event to the parent when I "mouseover" the child. That's not what I need. I need the "mouseout" event of the parent to not be sent when I "mouseover" the child. Get it? Event bubbling is useful when I don't want, for example, a click event on the child to be propagated to the parent, but this is not my case. What is weird is that I have other elements inside the same parent that don't fire the "mouseout" event of the parent when I "mouseover" them.
you can use "mouseenter" and "mouseleave" events available in jquery, here is the below code,
$(document).ready(function () {
$("#parent").mouseenter(function () {
$("#child").show();
});
$("#parent").mouseleave(function () {
$("#child").hide();
});
});
above is to attach an event,
<div id="parent">
<img id="child" src="button.png" style="display:none;" />
</div>
You can use the solution below, which it's pure javascript and I used with success.
var container = document.getElementById("container");
var mouse = {x: 0, y: 0};
function mouseTracking(e) {
mouse.x = e.clientX || e.pageX;
mouse.y = e.clientY || e.pageY;
var containerRectangle = container.getBoundingClientRect();
if (mouse.x > containerRectangle.left && mouse.x < containerRectangle.right &&
mouse.y > containerRectangle.top && mouse.y < containerRectangle.bottom) {
// Mouse is inside container.
} else {
// Mouse is outside container.
}
}
document.onmousemove = function () {
if (document.addEventListener) {
document.addEventListener('mousemove', function (e) {
mouseTracking(e);
}, false);
} else if (document.attachEvent) {
// For IE8 and earlier versions.
mouseTracking(window.event);
}
}
I hope it helps.
Just add an if statement to your MyFuncOut function which checks the target is not the image
if (event.target !== imageNode) {
imageNode.style.display = 'none'
}
I've had problems with this too and couldn't get it to work. This is what I did instead. It's much easier and isn't JavaScript so it's faster and can't be turned off.
div.container {
opacity:0.0;
filter:alpha(opacity="00");
}
div.container:hover {
opacity:1.0;
filter:alpha(opacity="100");
}
You can even add the CSS3 transition for opacity to give it a smoother feel to it.
Simply set css of the child element pointer-event:none; See example:
#parent {
width: 500px;
height: 500px;
border: 1px solid black;
}
#child {
width: 200px;
pointer-events: none; # this does the trick
}
<div id="parent" onmouseover="MyFuncHover()" onmouseout="MyFuncOut()">
<img id="child" src="https://i2.wp.com/beebom.com/wp-content/uploads/2016/01/Reverse-Image-Search-Engines-Apps-And-Its-Uses-2016.jpg?w=640&ssl=1" />
</div>
<script type="text/javascript">
function MyFuncHover() {
console.log("mouse over")
}
function MyFuncOut() {
console.log("mouse out")
}
</script>
来源:https://stackoverflow.com/questions/10618001/javascript-mouseover-mouseout-issue-with-child-element