powershell multiple arguments in one variable

Question

I want to have variable number of args go into one variable script.ps1 first -second a -third b

what I want is to have "first -second a -third b" all be in one variable.

Is this possible without passing all the arguments as one string?

Considering you say you don't want a string, can you clarify what you are asking? Meaning are you trying to pass these and assign them to a single hash table object, assign them as params, pass them directly to a starting function (as arguments), or something else? — LinkBerest - SO sold our work
– LinkBerest - SO sold our work, Commented Jul 21, 2016 at 22:52
Are you looking for $args? What is the purpose of this? Why can't you simply use optional parameters? — Ansgar Wiechers
– Ansgar Wiechers, Commented Jul 22, 2016 at 13:34
I want to have one script be called (kind of acting as a parent) and then based on the first parameter blindly pass the rest of the arguments to another script. So I want the "-params" too — mccormickt12
– mccormickt12, Commented Jul 22, 2016 at 17:31

Community · Accepted Answer · 2020-06-20 09:12:55Z

5

You could use the ValueFromRemainingArguments option in the Parameter attribute of your one variable

script.ps1

param(
  [Parameter(Mandatory=$true,ValueFromRemainingArguments=$true)]
  [psobject[]]$InputObject
)

foreach($value in $InputObject)
{
  Write-Host $value
}

Now you can supply as many arguments as you want to:

PS C:\> .\script.ps1 many arguments can go here
many
arguments
can
go
here

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jul 21, 2016 at 23:12

Mathias R. Jessen

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Comments

Ansgar Wiechers · Accepted Answer · 2016-07-22 21:44:28Z

1

If you want to treat the first argument in the list as a subcommand and pass the rest of the arguments on to other functions/cmdlets I'd simply split'n'splat the automatic variable $args, e.g. like this:

$subcommand, $params = $args

switch ($subcommand) {
  'foo'   { Do-Some @params }
  'bar'   { Do-Other @params }
  default { Write-Error "Unknown subcommand: $_" }
}

answered Jul 22, 2016 at 21:44

Ansgar Wiechers

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