Question

我试图在Powershell中抛出一个无控制窗口，当脚本仍然在执行时，它停留在顶部。一旦剩下的代码完成，我想删除无控制窗口。

除了如何移除窗口本身之外，我已经完成了所有工作。 $ Form.close（）不起作用，如果我不得不猜测它，因为它无法在函数中访问$ Form。

所以我不知所措。如何删除窗口？

function goForm
{
Add-Type -AssemblyName System.Windows.Forms
$Label = New-Object System.Windows.Forms.Label
$Label.AutoSize = $True
$Label.Text = "This form is very simple.  
In some cases, it will be larger.

And have spaces.

Like this."
$Form = New-Object System.Windows.Forms.Form
$Form.Text = "Progress"
$Form.SizeGripStyle = "Hide"
$Form.AutoScroll = $True
$Form.AutoSize = $True
$Form.AutoSizeMode = "GrowAndShrink"
$Form.MinimizeBox = $False
$Form.MaximizeBox = $False
$Form.Controlbox = $False
$Form.WindowState = "Normal"
$Font = New-Object System.Drawing.Font("Times New Roman",12,   [System.Drawing.FontStyle]::Regular)
$Form.Font = $Font
$Form.StartPosition = "CenterScreen" 
$Form.Controls.Add($Label)
$Form.Topmost = $True
$Form.Add_Shown({$Form.Activate()})
$Form.ShowDialog()
}

start-job $function:goForm

Start-Sleep -m 5000

$Form.Close()

[System.Reflection.Assembly]::LoadWithPartialName("System.Windows.Forms")
$oReturn=[System.Windows.Forms.Messagebox]::Show("The other window should be gone." , "")

Answer 1

如果使用Start-Job，则表单将在另一个PowerShell进程中创建，因此您无法在原始PowerShell进程中按变量访问表单。但是，您不需要显示此模式，因此您不需要使用Start-Job。只需使用.Show()代替，例如：

Add-Type -AssemblyName System.Windows.Forms
$Label = New-Object System.Windows.Forms.Label
$Label.AutoSize = $True
$Label.Text = @"
This form is very simple.
In some cases, it will be larger.

And have spaces.

Like this.
"@
$Form = New-Object System.Windows.Forms.Form
$Form.Text = "Progress"
$Form.SizeGripStyle = "Hide"
$Form.AutoScroll = $True
$Form.AutoSize = $True
$Form.AutoSizeMode = "GrowAndShrink"
$Form.MinimizeBox = $False
$Form.MaximizeBox = $False
$Form.Controlbox = $False
$Form.WindowState = "Normal"
$Font = New-Object System.Drawing.Font("Times New Roman",12,   [System.Drawing.FontStyle]::Regular)
$Form.Font = $Font
$Form.StartPosition = "CenterScreen"
$Form.Controls.Add($Label)
$Form.Topmost = $True
$Form.Add_Shown({$Form.Activate()})

$Form.Show()
$Form.Close()

使用功能时，无法关闭Powershell中的Form

1 个答案: