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

问题描述

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

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

如果运行以下命令，则无输出:

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文件，则会得到预期的结果:

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

这就是

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

这篇关于使用启动过程捕获标准输出和错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！