无法获取弹出窗口的URL /位置

Question

Is there a way to fetch the URL / Location of the pop-up window? 有没有一种方法可以获取弹出窗口的URL /位置？

CODE: 码：

<html>
   <head>
   <script>
   function openWin()
   {
      myWindow=window.open('http://www.google.com','','width=200,height=100');
      console.debug(myWindow.location.href);
      console.debug(window.location.href);
   }
   </script>
  </head>

  <body>
    <input type="button" value="Open window" onclick="openWin()" />
  </body>
</html>

The first console prints about:blank 第一个控制台显示以下about:blank

While the second console prints the URL of the current page(ie URL of the above code, it does not print the URL of the pop-up window) 当第二个控制台打印当前页面的URL（即上面代码的URL时，它不打印弹出窗口的URL）

Why isn't the first console printing the location (ie http://www.google.com ) ? 为什么第一个控制台不打印位置（即http://www.google.com ）？

Can anyone help me out with this? 谁能帮我这个忙吗？

Thanks in advance. 提前致谢。

Answer 1

As @Hg3 says, you cannot access location.href for myWindow . 正如@ Hg3所说，您无法访问myWindow location.href 。 window.open returns an empty DOMWindow . window.open返回一个空的DOMWindow 。

However, you can overwrite window.open and maintain a simple array of all the windows you have opened. 但是，您可以覆盖window.open并维护所有已打开窗口的简单数组。

//override window.open
var windows = [];
window._open = window.open;
window.open = function(url, name, params){
    windows.push(name, url);
    return window._open(url, name, params)
}
//function return href by name
function hrefByName(name) {
    var index=(windows.indexOf(name));
    return (index>-1) ? windows[index+1] : 'undefined';
}

//modified openWin function
function openWin(url, name) {
    var params='width=200,height=100';
    window.open(url, name, params);
    //test, ouput current url and the windows array
    console.log(hrefByName(name));
    console.log(windows);
}

test markup : 测试标记：

<input type="button" value="google" onclick="openWin('http://google.com', 'google')" />
<input type="button" value="bing" onclick="openWin('http://bing.com', 'bing')" />
<input type="button" value="stackoverflow" onclick="openWin('http://stackoverflow.com', 'stackoverflow')" />

Of course, I guess you have a setup with dynamically generated URL's - just build random names or pass an unique number as name. 当然，我想您有一个带有动态生成的URL的设置-只需构建随机名称或传递唯一数字作为名称即可。

Answer 2

Because browser security prevents you from getting the URL from any window that's not on the same domain as your script. 因为浏览器安全性阻止您从与脚本不在同一域中的任何窗口获取URL。 So, if your code was running on example.com, you'd only be able to get the URL of any window that was also on example.com. 因此，如果您的代码在example.com上运行，则只能获取也在example.com上的任何窗口的URL。

无法获取弹出窗口的URL /位置

问题描述

2 个解决方案

解决方案1
1 2013-06-04 14:12:33

解决方案2
0 2013-06-04 13:51:12

无法获取弹出窗口的URL /位置

问题描述

2 个解决方案

解决方案1 1 2013-06-04 14:12:33

解决方案2 0 2013-06-04 13:51:12

解决方案1
1 2013-06-04 14:12:33

解决方案2
0 2013-06-04 13:51:12