PowerShell将命名参数传递给ArgumentList

Question

我有一个PowerShell脚本,它接受3个命名参数.请让我知道如何从命令行传递相同的内容.我尝试下面的代码,但同样不起作用.它仅将整个值分配给P3.我的要求是P1应该包含1,P2应该包含2,而P3应该包含3.

Invoke-Command -ComputerName server -FilePath "D:\test.ps1" -ArgumentList  {-P1 1 -P2 2 -P3 3}

以下是脚本文件代码.

Param (
    [string]$P3,
    [string]$P2,
    [string]$P1
)
Write-Output "P1 Value :" $P1
Write-Output "P2 Value:" $P2
Write-Output "P3 Value :" $P3

Answer 1

一种选择:

$params = @{
P1 = 1
P2 = 2 
P3 = 3
}

$ScriptPath = 'D:\Test.ps1'

$sb = [scriptblock]::create(".{$(get-content $ScriptPath -Raw)} $(&{$args} @params)")

Invoke-Command -ComputerName server -ScriptBlock $sb

Answer 2

mjolinor的代码效果很好，但是花了我几分钟时间才理解。

该代码使事情变得简单-用内置参数生成脚本块的内容：

&{
    Param (
        [string]$P3,
        [string]$P2,
        [string]$P1
    )
    Write-Output "P1 Value:" $P1
    Write-Output "P2 Value:" $P2
    Write-Output "P3 Value:" $P3
} -P1 1 -P2 2 -P3 3

然后，此脚本块将传递给Invoke-Command。

为了简化代码：

".{$(get-content $ScriptPath -Raw)} $(&{$args} @params)"

$scriptContent = Get-Content $ScriptPath -Raw
$formattedParams = &{ $args } @params
# The `.{}` statement could be replaced with `&{}` here, because we don't need to persist variables after script call.
$scriptBlockContent = ".{ $scriptContent } $formattedParams"
$sb = [scriptblock]::create($scriptBlockContent)

让我们做一个基本的C＃实现：

void Run()
{
    var parameters = new Dictionary<string, string>
    {
        ["P1"] = "1",
        ["P2"] = "2",
        ["P3"] = "3"
    };

    var scriptResult = InvokeScript("Test.ps1", "server", parameters)
    Console.WriteLine(scriptResult);
}

string InvokeScript(string filePath, string computerName, Dictionary<string, string> parameters)
{
    var innerScriptContent = File.ReadAllText(filePath);
    var formattedParams = string.Join(" ", parameters.Select(p => $"-{p.Key} {p.Value}"));
    var scriptContent = "$sb = { &{ " + innerScriptContent + " } " + formattedParams + " }\n" +
        $"Invoke-Command -ComputerName {computerName} -ScriptBlock $sb";

    var tempFile = Path.Combine(Path.GetTempPath(), Guid.NewGuid() + ".ps1");
    File.WriteAllText(tempFile, scriptContent);

    var psi = new ProcessStartInfo
        {
            FileName = "powershell",
            Arguments = $@"-ExecutionPolicy Bypass -File ""{tempFile}""",
            RedirectStandardOutput = true,
            UseShellExecute = false
        };

    var process = Process.Start(psi);
    var responseText = process.StandardOutput.ReadToEnd();

    File.Delete(tempFile);

    return responseText;
}

该代码生成一个临时脚本并执行它。

示例脚本：

$sb = {
    &{
        Param (
            [string]$P3,
            [string]$P2,
            [string]$P1
        )
        Write-Output "P1 Value:" $P1
        Write-Output "P2 Value:" $P2
        Write-Output "P3 Value:" $P3
     } -P1 1 -P2 2 -P3 3
}
Invoke-Command -ComputerName server -ScriptBlock $sb