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Call two events on mouseover?

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Aug 07.2006

I’d like to have two seperate images have a mouseover effect when only one of them has someone mouse over them.

In other words, I have two images, and when either one of them someone mouses over them, there will be a mouseover effect for BOTH simultaneously.

is this possible?

thanks

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JavaScript

4 Comments(s) ↴

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@mark_yiehAug 07.2006 — firstImage.onmouseover = function() {

firstImageEffect();

secondImage Effect();

};

secondImage.onmouseover = function() {

firstImageEffect();

secondImageEffect();

};

function firstImageEffect() {

create the effect for the first image in this function.

}

function secondImageEffect() {

create the effect for the second image in this fucntion.

}
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@blackgoldauthorAug 07.2006 — firstImage.onmouseover = function() {

firstImageEffect();

secondImage Effect();

};

secondImage.onmouseover = function() {

firstImageEffect();

secondImageEffect();

};

function firstImageEffect() {

create the effect for the first image in this function.

}

function secondImageEffect() {

create the effect for the second image in this fucntion.

}[/QUOTE]

thanks! let me give it a shot ?
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@mark_yiehAug 08.2006 — you can make it even simpler by setting up a function that will take in the image as a variable and add the effect to the image. So instead of having two functions firstImageEffect, and secondImageEffect, you can create a function like this:

function imageEffect(targetImage) {

create the effect to targetImage in this function.

}

Then you can just call that function twice, each with a different variable like this:

firstImage.onmouseover = function() {

imageEffect(firstImage);

imageEffect(secondImage);

};

secondImage.onmouseover = function() {

imageEffect(firstImage);

imageEffect(secondImage);

};
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@konithomimoAug 08.2006 — Just create the two seperate functions and then add the following to both images:

onmouseover="function1();funtion2();"

where function1 and function2 are replaced by the names of your two functions.
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