使用 Start-Process 捕获标准输出和错误 [英] Capturing standard out and error with Start-Process

问题描述

在访问 StandardError 和 StandardOutput 属性时，PowerShell 的 Start-Process 命令是否存在错误?

Is there a bug in PowerShell's Start-Process command when accessing the StandardError and StandardOutput properties?

如果我运行以下命令，则没有输出:

If I run the following I get no output:

$process = Start-Process -FilePath ping -ArgumentList localhost -NoNewWindow -PassThru -Wait
$process.StandardOutput
$process.StandardError

但是如果我将输出重定向到一个文件，我会得到预期的结果:

But if I redirect the output to a file I get the expected result:

$process = Start-Process -FilePath ping -ArgumentList localhost -NoNewWindow -PassThru -Wait -RedirectStandardOutput stdout.txt -RedirectStandardError stderr.txt

推荐答案

Start-Process 是出于某种原因设计的.这是一种无需发送到文件即可获取它的方法:

That's how Start-Process was designed for some reason. Here's a way to get it without sending to file:

$pinfo = New-Object System.Diagnostics.ProcessStartInfo
$pinfo.FileName = "ping.exe"
$pinfo.RedirectStandardError = $true
$pinfo.RedirectStandardOutput = $true
$pinfo.UseShellExecute = $false
$pinfo.Arguments = "localhost"
$p = New-Object System.Diagnostics.Process
$p.StartInfo = $pinfo
$p.Start() | Out-Null
$p.WaitForExit()
$stdout = $p.StandardOutput.ReadToEnd()
$stderr = $p.StandardError.ReadToEnd()
Write-Host "stdout: $stdout"
Write-Host "stderr: $stderr"
Write-Host "exit code: " + $p.ExitCode

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