javascript - 如何打开一个弹出窗口，使其高度从屏幕顶部到底部？

Question

我试图打开一个弹出窗口，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？ (在 Chrome 中工作)

谢谢

Answer 1

我认为您只需要 screen.width 和 screen.height。不要在它们前面加上窗口。

编辑:显然 IE 需要一个“fullscreen=yes”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>