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？（在chrome中工作）

谢谢

Answer 1

我认为你只需要screen.width和screen.height。不要在窗口前面添加它们。

编辑：显然IE需要“全屏=是”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>

Answer 2

这对我有用：

<a class="popme" href="http://codesheet.org">popup</a>

的jQuery

  $(document).ready(function() {
    $(".popme").click(function(event){
      var h = 650;
      var w = 340;
      var wh = screen.height;
      var ww = screen.width;
      var top = wh/2 - h/2;
      var left = ww/2 - w/2;
      var popup = window.open(this.href + '', 'player', 'height=' + wh + ', width=' + w + ', scrollbars=no, left=' + left + ', top=' + top);
      popup.focus();
      return false;
    });
  });

演示：http://codesheet.org/codesheet/JMbOYCUI

Answer 3

如果您只想全屏打开新窗口，是否尝试删除高度和宽度参数？

newWindow = window.open(windowUri, windowName);

see reference

如何打开弹出窗口，使其高度从屏幕顶部到底部？

3 个答案: