Question

在powershell中，我可以使用

启动一个进程

$app_name = "app.exe"
$app_arguments = "arg0"
Start-Process $app_name $app_arguments

我尝试用

设置亲和力

$app = Start-Process $app_name $app_arguments
$app.ProcessorAffinity = 0x3

....没有工作。

在windows powershell中，当我启动一个进程时如何设置亲和力？

Answer 1

我可以用

解决

$app_name = "app.exe"
$app_arguments = "arg0"

$pinfo = New-Object System.Diagnostics.ProcessStartInfo
$pinfo.FileName = $app_name
$pinfo.Arguments = $app_arguments
$p = New-Object System.Diagnostics.Process
$p.StartInfo = $pinfo
$p.Start()
$p.ProcessorAffinity=0x3

Answer 2

PowerShell启动脚本

我错过了DOS start命令，因此我将@ JuanPablo的代码合并到一个名为PSstart.ps1的shell脚本中，您可以使用它替换PowerShell中的start命令。

只需使用 PowerShell -file PSStart.ps1 -affinity <affinity> -priority <priority> <path to executable> <executable arguments> 享受吧！

param([Int32]$affinity=0xF,[String]$priority="NORMAL", [String]$appPath="", [String]$appArguments="")
[System.Reflection.Assembly]::LoadWithPartialName("System.Windows.Forms")   # For message box error reports, remove if you don't want popup errors

$priorityValues = "LOW", "NORMAL", "HIGH", "REALTIME", "ABOVENORMAL", "BELOWNORMAL" # Remove ABOVENORMAL and BELOWNORMAL if running on Win98 or WinME
$priorityUC = $priority.ToUpper()
$pinfo = New-Object System.Diagnostics.ProcessStartInfo

If($appPath -ne "" -and (Test-Path $appPath))
{
    If($priorityValues -contains $priorityUC)
    {
        Try
        {
            $pinfo.FileName = $appPath
            $pinfo.Arguments = $app_arguments
            $p = New-Object System.Diagnostics.Process
            $p.StartInfo = $pinfo
            $p.Start()
            $p.PriorityClass=$priorityUC
            $p.ProcessorAffinity=$affinity
        }
        Catch
        {
            $exceptionMessage = $_.Exception.Message
            #Write-Host "An exception:`n`n$exceptionMessage`n`noccured!" -fore white -back red  # Uncomment for console errors
            [System.Windows.Forms.MessageBox]::Show("An exception:`n`n$exceptionMessage`n`noccured!", "An Exception Occured", "Ok", "Error");
            Break
        }
    }
    Else
    {
        #Write-Host "The priority: `"$priorityUC`" is not a valid priority value!" -fore white -back red    # Uncomment for console errors
        [System.Windows.Forms.MessageBox]::Show("The priority: `"$priorityUC`" is not a valid priority value!", "A Priority Error Occured", "Ok", "Error");
    }
}
Else
{
    #Write-Host "The application path: `"$appPath`" doesn't exist!", "A Path Error Occured" -fore white -back red   # Uncomment for console errors
    [System.Windows.Forms.MessageBox]::Show("The application path: `"$appPath`" doesn't exist!", "A Path Error Occured", "Ok", "Error");
}

Answer 3

您需要传递-PassThru开关才能获取过程对象

$app = Start-Process $app_name $app_arguments -PassThru
$app.ProcessorAffinity = 0x3

根据powershell Start-Process command (from ps 3.0)

-PassThru为cmdlet启动的每个进程返回一个进程对象。默认情况下，此cmdlet不会生成任何输出。

在powershell中，在启动过程中设置关联

3 个答案:

PowerShell启动脚本