mouseout stopping a function

the other john · May 12, 2006

I have a function that is activated by a mouseover. The function
triggers an image rotation. I need to stop the rotation on the
mouseout but I don't know how to do this. the mouseover triggers the
rotate() function below. currently the mouseout produces the default
image but then it keeps cycling the other images.

<script language="javascript" type="text/javascript">
if(document.images) {
bubbles = new Image
off = new Image
bubbles.src = "images/bubbles.jpg"
off.src = "default.jpg"
}
else {
bubbles = ""
off = ""
}

adImages = new Array("images/whitemarble.jpg", "images/bubbles.jpg",
"images/oak.jpg")
thisAd = 0

imgCt = adImages.length

function rotate() {
if (document.images) {
thisAd++
if(thisAd == imgCt) {
thisAd = 0
}
document.ImgField.src=adImages[thisAd]
setTimeout("rotate()", 3 * 1000)
}
}
</script>

Evertjan. · May 12, 2006

the other john wrote on 12 mei 2006 in comp.lang.javascript:

I have a function that is activated by a mouseover. The function
triggers an image rotation. I need to stop the rotation on the
mouseout but I don't know how to do this. the mouseover triggers the
rotate() function below. currently the mouseout produces the default
image but then it keeps cycling the other images.

<script language="javascript" type="text/javascript">

Do not use language="javascript" anymore

if(document.images) {
bubbles = new Image
off = new Image
bubbles.src = "images/bubbles.jpg"
off.src = "default.jpg"
}
else {
bubbles = ""
off = ""
}

I don't know why you need the above code.
After all you are not preloading all images you use.

adImages = new Array("images/whitemarble.jpg", "images/bubbles.jpg",
"images/oak.jpg")
thisAd = 0

imgCt = adImages.length

use var in defining new variables.

var imgCt = adImages.length;
var myInterval;
etc.
var ImgField = document.getElementById('ImgField'); //see below

function rotate() {
if (document.images) {
thisAd++
if(thisAd == imgCt) {
thisAd = 0
}

You will see see number 0 last this way

document.ImgField.src=adImages[thisAd]

Use ID and getElementById() bor a better cross-browser compatibility:

ImgField.src=adImages[thisAd]

delete the below line:

setTimeout("rotate()", 3 * 1000)
}
}
</script>

and do:

onmouseover = 'myInterval = setInterval("rotate()", 3e3);'
onmouseout =
'clearInterval(myInterval);imgField.src="images/default.jpg"'

==============

Try in the <head>:

<script type='text/javascript'>
var adImages =
['images/whitemarble.jpg','images/bubbles.jpg','images/oak.jpg']
var imgCt = adImages.length;
var imageHolder = new Array;

for (var i=0;i<imgCt;i++){ // preloader
imageHolder = new Image()
imageHolder.src = adImages
}

var thisAd = 0;
var myInterval;

function rotate() {
document.getElementById('ImgField').src =
imageHolder[thisAd].src
if (++thisAd == imgCt) thisAd = 0
}
</script>

[not tested]

the other john · May 12, 2006

That worked, excellent!

Now I have to figure out how to make this work for 12 different sets of
rollovers. I know there's a way to do it with a more involved array
but I"m not that experienced.

There are 12 thumbs in a circle. the ImgField is in the center. when
one of the thumbs in the circle is moused over the center image,
ImgField, displays a rotation of samples of the thumb being rolled
over. mouseout produces the default image. What you gave me works
great! if I can make it for all 12 without writing a seperate function
for each thumb that would be great.

Thanks for your help! I might be able to feed my kids this week ;-)

Evertjan. · May 12, 2006

the other john wrote on 12 mei 2006 in comp.lang.javascript:

That worked, excellent!

What worked?

Please quote what you are replying to.

If you want to post a followup via groups.google.com, don't use the
"Reply" link at the bottom of the article. Click on "show options" at the
top of the article, then click on the "Reply" at the bottom of the article
headers. <http://www.safalra.com/special/googlegroupsreply/>

the other john · May 12, 2006

this worked....arg...posting on google is a pain....

the other john wrote on 12 mei 2006 in comp.lang.javascript:

I have a function that is activated by a mouseover. The function
triggers an image rotation. I need to stop the rotation on the
mouseout but I don't know how to do this. the mouseover triggers the
rotate() function below. currently the mouseout produces the default
image but then it keeps cycling the other images.

<script language="javascript" type="text/javascript">

Do not use language="javascript" anymore

if(document.images) {
bubbles = new Image
off = new Image
bubbles.src = "images/bubbles.jpg"
off.src = "default.jpg"
}
else {
bubbles = ""
off = ""
}

I don't know why you need the above code.
After all you are not preloading all images you use.

adImages = new Array("images/whitemarble.jpg", "images/bubbles.jpg",
"images/oak.jpg")
thisAd = 0

imgCt = adImages.length

use var in defining new variables.

var imgCt = adImages.length;
var myInterval;
etc.
var ImgField = document.getElementById('ImgField'); //see below

function rotate() {
if (document.images) {
thisAd++
if(thisAd == imgCt) {
thisAd = 0
}

You will see see number 0 last this way

document.ImgField.src=adImages[thisAd]

Use ID and getElementById() bor a better cross-browser compatibility:

ImgField.src=adImages[thisAd]

delete the below line:

setTimeout("rotate()", 3 * 1000)
}
}
</script>

and do:

onmouseover = 'myInterval = setInterval("rotate()", 3e3);'
onmouseout =
'clearInterval(myInterval);imgField.src="images/default.jpg"'

==============

Try in the <head>:

<script type='text/javascript'>
var adImages =
['images/whitemarble.jpg','images/bubbles.jpg','images/oak.jpg']
var imgCt = adImages.length;
var imageHolder = new Array;

for (var i=0;i<imgCt;i++){ // preloader
imageHolder = new Image()
imageHolder.src = adImages

}

var thisAd = 0;
var myInterval;

function rotate() {
document.getElementById('ImgField').src =
imageHolder[thisAd].src
if (++thisAd == imgCt) thisAd = 0

}

</script>

[not tested]

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mouseout stopping a function

the other john

Evertjan.

the other john

Evertjan.

the other john

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