Powershell parameter passing to function

Question

I'm facing with an annoying problem: my script seemingly doesn't pass any argument to a function I've defined.

$server = 'http://127.0.0.1:8080' Function Get-WorkingDirectory([string]$address) { #echo $address $content = Get-Content -path C:\....\file.txt $content -contains $address } #end Get-WorkingDirectory function if(Get-WorkingDirectory $server) { echo "works" } else { echo "error" }

It is stuck on "works". If I try to echo address in the function, it is empty. What the heck I'm doing wrong?! I know this is a pretty noobish question, but I tried everything I found on the net. Thanks in advance for help!

@MathiasR.Jessen the if condition is true, then it echoes "works", also if there is no evaluation — Kernel
– Kernel, Commented Apr 27, 2015 at 13:26

Lieven Keersmaekers · Accepted Answer · 2015-04-27 11:44:20Z

echo is an alias for Write-Output but as you are using the output of the function in the if statement, nothing gets shown.

For testing purposes, use Write-Host in this instance to show the variable being passed correctly.

$server = 'http://127.0.0.1:8080' Function Get-WorkingDirectory([string]$address) { write-host "$address using write host" } #end Get-WorkingDirectory function if (Get-WorkingDirectory $server) { }

ДМИТРИЙ МАЛИКОВ · Accepted Answer · 2015-04-27 11:41:03Z

Output of Get-WorkingDirectory is shadowed by if statement.

Try to use it without if and you'll see that argument is passed correctly. For example,

$server = 'http://127.0.0.1:8080' Function Get-WorkingDirectory([string]$address) { Write-Host $address } Get-WorkingDirectory $server

Address is printed well

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