powershell - Simple ScriptBlock 在本地工作，但不能在远程工作

Question

我有一个文件，我正在使用 PowerShell 4.0 从一台计算机传输到另一台计算机.我创建了一个读取缓冲区，将其转换为 Base64String，然后打开一个新的 PSSession。最后，我将这段代码称为:

#Open a file stream to destination path in remote session and write buffer to file
#First save string buffer to variable so FromBase64String() parses correctly in ScriptBlock
$remoteCommand = 
"#First save string buffer to variable so FromBase64String() parses correctly below
    `$writeString = `"$stringBuffer`"
    `$writeBuffer = [Convert]::FromBase64String(`"`$writeString`")
    `$writeStream = [IO.File]::Open(`"$destPath`", `"Append`")
    `$writeStream.Write(`$writeBuffer, 0, `$writeBuffer.Length)
    `$WriteStream.Close()
"
Invoke-Command -ScriptBlock{ Invoke-Expression $args[0] } -ArgumentList $remoteCommand -Session $remoteSession 

我试过运行

Invoke-Command -ScriptBlock{ Invoke-Expression $args[0] } -ArgumentList $remoteCommand

运行良好，创建文件并按预期写入 byte[]。当我跑

Invoke-Command -ScriptBlock{ Invoke-Expression $args[0] } -ArgumentList $remoteCommand -Session $remoteSession

我得到错误

Exception calling "Open" with "2" argument(s): "Could not find a part of the path 'C:\Test\3.txt'."

我期望这样做是解析远程机器上的命令，以便它在远程机器上创建一个新文件'C:\Test\3.txt'并附加字节[]。任何想法我怎么能做到这一点？

Answer 1

我错过了您通过 $stringBuffer 的部分到你的脚本 block 。但是，首先您可以使用大括号更轻松地编写脚本 block 。然后你可以使用 $using:VARIABLENAME传递本地脚本变量:

$remoteCommand = {
    #First save string buffer to variable so FromBase64String() parses correctly below
    $writeString = $using:stringBuffer
    $writeBuffer = [Convert]::FromBase64String($writeString)
    $writeStream = [IO.File]::Open($destPath, "Append")
    $writeStream.Write($writeBuffer, 0, $writeBuffer.Length)
    $WriteStream.Close()
}

Invoke-Command -ScriptBlock $remoteCommand -Session $remoteSession