在Windows Powershell中引用弹出窗口 [英] Reference to Pop-up window in Windows Powershell

问题描述

我正在为我正在研究的网站进行测试自动化。我使用Windows PowerShell来创建脚本来执行此操作。我的问题是我需要单击一个打开另一个窗口的链接，我需要一个如何引用该窗口。

I am working on testing-automation for a website I'm working on. I am using windows powershell to create scripts to do this. My problem is I need to click a link that opens up another window, I need a reference to that window some how.

$ie = new-object -com "InternetExplorer.Application"
$ie.navigate("http://localhost:4611")
$ie.visible = $true
$doc = $ie.document
$link = $doc.getElementByID('link')

这是我获得对浏览器和链接的引用的地方。然后我点击链接：

That is where I get a reference to the browser and the link. Then I click the link:

$link.click()

这会打开一个新窗口，其中包含我需要测试的内容。我如何获得对这个新窗口的引用？从技术上讲，它不是第一个的儿童窗口。

Which opens up a new window with things I need to test on it. How would I get a reference to this new window? It is not technically a child window of the first one.

我试图将链接的点击设置为这样的引用，但它不起作用：

I have tried to set the click of the link to a reference like this but it doesn't work:

$test = new-object -com "InternetExplorer.Application"
$test = $link.click()

更新：
这是调用打开窗口的JavaScript函数openwindow

UPDATE: Here is the JavaScript function openwindow that gets called to open the window

function openwindow(url, name) {
   if (typeof openwindow.winrefs == 'undefined') {
      openwindow.winrefs = {};
   }
   if (typeof openwindow.winrefs[name] == 'undefined' || openwindow.winrefs[name].closed) {
    openwindow.winrefs[name] = window.open(url, name, 'scrollbars=yes,menubar=no,height=515,width=700,resizable=no,toolbar=no,status=no');
   } else {
      openwindow.winrefs[name].focus();
   };
};

该函数在一行代码中调用，如下所示

The function gets called in a line of code that looks like this

column.For(i => "<a href='" + link + i.id + "?status=new' target='pat" + i.id + "'id'enc' onclick='openwindow(this.href,this.target);return false'>

最终更新：
我的结果略有不同。我创建了一个新的Internet Explorer对象并从链接中抓取了href并设置了所有选项并使用像javascript一样的powershell导航到窗口。

FINAL UPDATE: I ended up doing this slightly differently. I created a new Internet Explorer object and grabbed the href from the link and set all of the options and navigated to the window using powershell like the javascript does.

$ie2 = new-object -com "InternetExplorer"
$ie2.visible = $true
$ie2.menubar = $false
$ie2.height = 515
$ie2.width = 700
$ie2.resizable = $false
$link = $doc.getelementbyid('link')
$url = $link.href
$ie2.navigate($url)

我非常感谢@scunliffe帮助我解决这个问题。

I'd like to thank @scunliffe so much for helping me through this problem.

推荐答案

此方法中存在拼写错误：$ doc.getElementByID（'link'）

there's a typo in this method: $doc.getElementByID('link')

它应该是：

$doc.getElementById('link')
                  ^ lowercase d

更新：

根据跟进代码/评论...您应该能够使用以下内容提取对窗口的引用：

Based on the follow up code/comments... you should be able to extract a reference to the window with something like this:

$ie = new-object -com "InternetExplorer.Application"
$ie.navigate("http://localhost:4611")
$ie.visible = $true
$doc = $ie.document
$link = $doc.getElementById('link')
$link.click()

该链接具有一个目标属性集，openwindow函数使用该属性为弹出窗口指定名称。

The link has a target attribute set that the openwindow function uses to assign a name to the popup window.

openwindow函数本身存储对弹出窗口的引用在一个名为winrefs的属性中，因此现在应该得到窗口...（如果我对PowerShell sy的期望IE窗口中的ntax是正确的。

The openwindow function itself, stores a reference to the popup in a property called winrefs, thus this should now get the window... (if my expectations of the PowerShell syntax within the IE window are correct.

$targetName = $link.target
$popupHandle = $ie.openwindow.winrefs[$targetName]

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