如何打开弹出窗口,使其高度从屏幕顶部到底部？

Question

我正在尝试打开一个弹出窗口,使其高度从屏幕顶部到Windows的"应用程序"栏.这是代码:

function windowOpener(windowHeight, windowWidth, windowName, windowUri) {

var windowHeight = window.innerHeight ? window.innerHeight : document.documentElement.clientHeight ? document.documentElement.clientHeight : document.body.clientHeight;


var centerWidth = (window.screen.width - windowWidth) / 2;
var centerHeight = (window.screen.height - windowHeight) / 2;


newWindow = window.open(windowUri, windowName, 'resizable=0,scrollbars=1,width=' + windowWidth +
    ',height=' + windowHeight +
    ',left=' + centerWidth);

newWindow.focus();
return newWindow.name;

}

为什么它在IE中不起作用？(适用于镀铬)

谢谢

Answer 1

我想你只需要screen.width和screen.height.不要在窗口前面添加它们.

编辑:显然IE需要一个"全屏=是"的参数.

试试这个:

<script type="text/javascript">
<!--
function popup(url) 
{
 params  = 'width='+screen.width;
 params += ', height='+screen.height;
 params += ', top=0, left=0'
 params += ', fullscreen=yes';

 newwin=window.open(url,'windowname4', params);
 if (window.focus) {newwin.focus()}
 return false;
}
// -->
</script>

<a href="javascript: void(0)" 
   onclick="popup('popup.html')">Fullscreen popup window</a>