Use variable as -path when using Get-Content on PowerShell

Question

I am trying to get a content of a text file by using Get-Content and I want the value for the -path to be on a variable like so:

$MyFileName = "testfile"    
$MyFilePath = "(.\MyFolder\" + $MyFileName + ".txt)"
$ServerList = Get-Content -Path $MyFilePath

But I keep getting the error:

Cannot bind argument to parameter 'Path' because it is null.

It works if I hard code the file path

$ServerList = Get-Content -Path (.\MyFolder\MyFile.txt)

Write-Host $MyFilePath
.\MyFolder\testfile.txt

Don't use parentheses in the filename unless the files literally contain them. -Path expects a string or string array. In your case, that would be -Path '.\MyFolder\MyFile.txt'. — AdminOfThings
– AdminOfThings, Commented Jun 20, 2019 at 16:49
please, unless you NEED the parens around your Get-Content section, remove them. they are unusual enuf that they are distracting from the newly changed point of your question. if you do need them, please explain why. — Lee_Dailey
– Lee_Dailey, Commented Jun 20, 2019 at 17:27
I don't specifically need them, I am new at powershell and I basically come up with this code from various sources, i'll update my question once i get home — Ruben_PH
– Ruben_PH, Commented Jun 20, 2019 at 17:31

js2010 · Accepted Answer · 2019-06-20 18:35:23Z

2

If you look at the variable, the string literally has parentheses in it:

$MyFileName = "testfile"
$MyFilePath = "(.\MyFolder\" + $MyFileName + ".txt)"
$myfilepath

(.\MyFolder\testfile.txt)

This would work:

$MyFileName = "testfile"
$MyFilePath = ".\MyFolder\" + $MyFileName + ".txt"
$myfilepath

.\MyFolder\testfile.txt

You could put the parentheses on the outside, but you don't need to. Or

".\MyFolder\$MyFileName.txt"

edited Jun 20, 2019 at 18:35

answered Jun 20, 2019 at 17:51

js2010

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Comments

Guenther Schmitz · Accepted Answer · 2019-06-20 16:36:32Z

1

try setting the full file path like

$MyFilePath = "C:\My Folder\My File.txt"

or if you the relative path is really what you want remove the brackets like

$MyFilePath = ".\My Folder\My File.txt"

answered Jun 20, 2019 at 16:36

Guenther Schmitz

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6 Comments

Ruben_PH Over a year ago

tried it, i still get the same error. I edited my question because the actual filename is a variable in itself minus the file extension .txt

Guenther Schmitz Over a year ago

try using Join-Path like $MyFilePath = Join-Path -Path "C:\My Folder" -ChildPath $MyFileName if the filetype is missing add this line $MyFilePath += '.txt'

Ruben_PH Over a year ago

still the same, i even tried $MyFilePath = "C:\My Folder\{0}.txt" -f $MyFileName i get the correct result when I write-host $MyFilePath. The issue is when using the $MyFilePath variable for the -Path option on Get-Content

Guenther Schmitz Over a year ago

so what is $MyFileName really? what does $MyFileName | Get-Member return?

Ruben_PH Over a year ago

TypeName: System.String

|

Lee_Dailey · Accepted Answer · 2019-06-20 17:17:56Z

1

here's one way to do what you seem to want. [grin] the 1st part is your code with the very peculiar resulting file name. the 2nd part is broken out into parts that are easier to read/understand/modify.

$YourFileName = "testfile"    
$YourFilePath = "(.\MyFolder\" + $YourFileName + ".txt)"

$BaseName = 'testfile'
$Extension = 'txt'
$FileName = $FileName, $Extension -join '.'
$Directory = $env:TEMP
$FullFileName = Join-Path -Path $Directory -ChildPath $FileName

$YourFilePath
$FullFileName

output ...

(.\MyFolder\testfile.txt)
C:\Temp\testfile.txt

note that your code made a file name that is almost certainly invalid. [grin]

answered Jun 20, 2019 at 17:17

Lee_Dailey

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2 Comments

Ruben_PH Over a year ago

I think the issue stems from me using Get-Content inside a Workflow for parallel processing, when I take the code outside of the workflow, it worked flawlessly. I thought the issue was just with that block of code but I was mistaken, I will update my question.

Lee_Dailey Over a year ago

@Ruben_PH - ah! that changes things a good deal. as long as you fix that horribly broken file name maker code, folks can focus on other aspects. [grin]

Ruben_PH · Accepted Answer · 2019-06-21 07:19:54Z

I found out that the issue is on using a variable inside a workflow. I was so focused on that block of code that I forgot to look at the bigger picture.

I have the code below which had the issue:

Workflow GetServerStatus{
   $ServerList = Get-Content -path $FullFileName
   $ServiceList = Get-Content service_list.txt
   ForEach -Parallel ($Server in $ServerList){
      InlineScript{
         Get-Service -ComputerName $Using:Server -name $Using:ServiceList
      }
   }
}

#credits to @Lee_Dailey
$Extension = 'txt'
$FileName = $BaseName, $Extension -join '.'
$Directory = '.\server'
$FullFileName = Join-Path -Path $Directory -ChildPath $FileName

GetServiceStatus

It turns out that the issue is that I am not passing the variables correctly to a workflow It should be:

Workflow GetServiceStatus{
    param(
        $FullFileName
    )

It is then called like so

GetServiceStatus $FullFileName

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