Redirecting standard input\output in Windows PowerShell

Question

What is the required syntax to redirect standard input/output on Windows PowerShell?

On Unix, we use:

$./program <input.txt >output.txt 

How do I execute the same task in PowerShell?

Answer 1

You can't hook a file directly to stdin, but you can still access stdin.

Get-Content input.txt | ./program > output.txt 

Answer 2

If there is someone looking for 'Get-Content' alternative for large files (as me) you can use CMD in PowerShell:

cmd.exe /c ".\program < .\input.txt" 

Or you can use this PowerShell command:

Start-Process .\program.exe -RedirectStandardInput .\input.txt -NoNewWindow -Wait 

It will run the program synchronously in same window. But I was not able to find out how to write result from this command to a variable when I run it in PowerShell script because it always writes data to the console.

EDIT:

To get output from Start-Process you can use option

-RedirectStandardOutput

for redirecting output to file and then read it from file:

Start-Process ".\program.exe" -RedirectStandardInput ".\input.txt" -RedirectStandardOutput ".\temp.txt" -NoNewWindow -Wait $Result = Get-Content ".\temp.txt"