How to change the position of according to the screen resolution?

hulidd

I used the code of a floating window on the webpage.
When I adjusted the screen resolution to 1280 * 1024. It won't work. . Blocking the logo above.
I use
  if (window.screen.width == 1280)
   {
    document.write "<div id = 'topview' style = 'position: absolute; margin-left: -845px; width: 145px; top: 0px; z-index: 1;'>";
   }
  else
  {
    document.write "<div id = 'topview' style = 'position: absolute; margin-left: -1025px; width: 145px; top: 0px; z-index: 1;'>";
  }
This statement controls the position of the <div>. .
But it cannot be achieved. . .
js is not too familiar. . Is it a problem with the code? Or something else?

issalee

When you change the screen resolution, things on the webpage are not refreshed

I suggest you make the above code into a function
Use window.setInterval ("function to be called", 500) to call regularly

hulidd

Thanks for the reminder upstairs .. but the question is:
My code has a script error ... don't know why ...?
I haven't found anything wrong after searching for a long time ... but ...

chenfuyu

<table style='border:1px solid #000; width: 100px; height: 200px;' align=right>
<tr><td>sdafsadf</td></tr></table>

Set near right margin

chenfuyu

<div style='border:1px solid #000; width: 100px; height: 200px; position:absolute;right:1px'>sdafsadf</div>

liub950511

Use relative position, just position it relative to your logo.

code show as below:

function GetRelated(element,attribute)
{
The
    if (arguments.length != 2 || element == null)
    {
        return null;
    }

    var elmt = element;

    var offsetTop = elmt.offsetTop;

var offsetLeft = elmt.offsetLeft;
The
    while( elmt.offsetParent) {

elmt = elmt.offsetParent

        offsetTop += elmt.offsetTop;

offsetLeft += elmt.offsetLeft;

    }

if( attribute =='top') return offsetTop;

else if( attribute =='left') return offsetLeft;

else return 0;

}