使用PowerShell(Start-Process)启动进程时是否可以定位窗口？

Question

我按如下方式运行命令.

Start-Process dotnet -ArgumentList "run"

可以使用-WindowStyle标记来管理窗口以使其最大化,最小化,隐藏和正常.但是,我通常做的是将框架向左推(和第二个向右).

是否有可能告诉PowerShell将窗口浮动到边缘？像这个如意的伪代码？

Start-Process dotnet -ArgumentList "run" -WindowStyle FloatLeft

Answer 1

试试这个，它使用 的-Passthru选项来Start-Process获取进程信息。然后，我们使用一些pInvoke魔法将刚刚创建的窗口移动到其他位置。

此示例将使您能够将生成的窗口捕捉到窗口当前屏幕的边缘。您可以指定 X 或 Y 边缘，或两者都指定。如果指定了所有 4 个开关，则顶部、左侧获胜。

Add-Type -AssemblyName System.Windows.Forms

Add-Type @"
using System;
using System.Collections;
using System.Collections.Generic;
using System.Runtime.InteropServices;
using System.Text;

public struct RECT
{
    public int left;
    public int top;
    public int right;
    public int bottom;
}

public class pInvoke
{
    [DllImport("user32.dll", SetLastError = true)]
    public static extern bool MoveWindow(IntPtr hWnd, int X, int Y, int nWidth, int nHeight, bool bRepaint);

    [DllImport("user32.dll", CharSet = CharSet.Auto, CallingConvention = CallingConvention.StdCall, ExactSpelling = true, SetLastError = true)]
    public static extern bool GetWindowRect(IntPtr hWnd, ref RECT rect);
}
"@

function Move-Window([System.IntPtr]$WindowHandle, [switch]$Top, [switch]$Bottom, [switch]$Left, [switch]$Right) {
  # get the window bounds
  $rect = New-Object RECT
  [pInvoke]::GetWindowRect($WindowHandle, [ref]$rect)

  # get which screen the app has been spawned into
  $activeScreen = [System.Windows.Forms.Screen]::FromHandle($WindowHandle).Bounds

  if ($Top) { # if top used, snap to top of screen
    $posY = $activeScreen.Top
  } elseif ($Bottom) { # if bottom used, snap to bottom of screen
    $posY = $activeScreen.Bottom - ($rect.bottom - $rect.top)
  } else { # if neither, snap to current position of the window
    $posY = $rect.top
  }

  if ($Left) { # if left used, snap to left of screen
    $posX = $activeScreen.Left
  } elseif ($Right) { # if right used, snap to right of screen
    $posX = $activeScreen.Right - ($rect.right - $rect.left)
  } else { # if neither, snap to current position of the window
    $posX = $rect.left
  }

  [pInvoke]::MoveWindow($app.MainWindowHandle, $posX, $posY, $rect.right - $rect.left, $rect.bottom - $rect.top, $true)
}

# spawn the window and return the window object
$app = Start-Process dotnet -ArgumentList "run" -PassThru

Move-Window -WindowHandle $app.MainWindowHandle -Bottom -Left