Powershell弹出窗口关闭应用程序

Question

我想创建一个powershell文件，该文件将创建一个弹出窗口，警告某人应用程序将在15分钟内自动关闭，除非他们延长会话。然后它应该终止应用程序，除非它们扩展会话。

我对此完全陌生，这就是我用记事本应用程序尝试的全部内容。

Add-Type -AssemblyName PresentationCore,PresentationFramework
$ButtonType = [System.Windows.MessageBoxButton]::YesNoCancel
$MessageIcon = [System.Windows.MessageBoxImage]::Error
$MessageBody = "Can we close notepad?"
$MessageTitle = "Using Notepad"
$Result = [System.Windows.MessageBox]::Show($MessageBody, $MessageTitle, $ButtonType, $MessageIcon)
Write-Host "Your choice is $Result"
Get-Process Notepad |   Foreach-Object { $_.CloseMainWindow() | Out-Null }

Answer 1

Mathias，如果您将其更改为一个开关，它的工作方式就像您想要的一样。关闭窗口需要Yes，但取消则保持打开

Add-Type -AssemblyName PresentationCore,PresentationFramework
$ButtonType = [System.Windows.MessageBoxButton]::YesNoCancel
$MessageIcon = [System.Windows.MessageBoxImage]::Error
$MessageBody = "Can we close notepad?"
$MessageTitle = "Using Notepad"
$Result = [System.Windows.MessageBox]::Show($MessageBody, $MessageTitle, $ButtonType, $MessageIcon)
switch($Result){
"Yes" {
write-host "You selected $Result"
Get-Process notepad |   Foreach-Object { $_.CloseMainWindow() | Out-Null } | Stop-Process -Force
}
"No" {
Write-Host "You selected $Result"
} 
"Cancel" {
Write-Host "You selected $Result"
} 
}