Screen size or browser window size??

[bewin99]

The following code apparently measures the screen size and not the browser window size, which is the wrong way to go about testing screen suitability for a website.

I have designed three sites with the following dimensions as a basis: 640x480, 800x600 and 1024x768 and was sure that screen size was right ~is it?

var width = screen.width;var res =(((!(640-width))*1)+((!(800-width))*2)+((!(1024-width))*3)+((!(1152-width))*4)+((!(1280-width))*5)+((!(1600-width))*6));
if(!(res)) res = 1;if (res=='1') {window.location='

http://www.640.com/640_480.html'}

if (res=='2') {window.location='

http://www.86.com/800_600.html'}

if (res=='3') {window.location='

http://www.1024+.com/1024_768.html'}

if (res!='1' && res!='2' && res!='3') PopupCenter ('

http://www.meta.projectmio.com/800.html','MyWindow',800,531)

}

 

[vicvirk]

you don't need to do any calculations to determine the screen width...use "screen.width" to determine it and place your actions within a switch:

<script language="javascript" type="text/javascript">
switch (screen.width) {
	case 640:
		// do 640 x 480 task
		break;
	case 800:
		// do 800 x 600 task
		break;
	case 1024:
		// do 1024 x 768 task
		break;
	default:
		// do default task (none of the above)
		break;
}
</script>

--------

GOOGLE is a great resource to find answers to questions like "how do i..."

http://www.google.com/webhp?complete=1&hl=en

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[BillyRayPreachersSon]

Also, think whether you really need to use JavaScript to do this, or whether coding a fluid layout using CSS alone would be a better (and nicer) option.

Dan



Coedit Limited - Delivering standards compliant, accessible web solutions

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