How to show Get-Help in powershell when no parameters are given

Question

Ok, I tried to look around and figure out how to do this, but wasn't able to figure it out. Here's what I want to do.

# Run script normally
.\myscript.ps1 "blah" "yo" 

These should do the same thing

Get-Help .\myscript.ps1
.\myscript.ps1

I want to be able to use the Get-Help syntax, but don't want to have to repeat it somewhere with a parameter switch statement. If this has been answered somewhere else, feel free to point me there.

Answer 1

You can add the following code to the top of your script:

if($args.Count -eq 0) {
    Get-Help $MyInvocation.MyCommand.Definition
    return
}

Answer 2

$PSBoundParameters.Values.Count will give you the count of parameters passed when you use param() in your script. You can then use $args to find anything else entered that is outside these parameters. So if you test for both when using param you can get help response.

Example parametertest.ps1

param([string]$val1,$val2,[switch]$val3)
if ( $PSBoundParameters.Values.Count -eq 0 -and $args.count -eq 0 ) {
    Get-Help $MyInvocation.MyCommand.Definition
    return 
    }
if ( $PSBoundParameters.Values.Count -eq 0 ){ 
    Write-Output ("No parameters passed")     
    return
    }
else {     write-output("val1=$val1 val2=$val2 val3=$val3") }
if ( $args -and $args.count -gt 0 ) {
    Write-Output ("Found Args $args")  }
if ( $PSBoundParameters.Values.Count -eq 0 -and  $args -and $args.count -eq 0

Example output

powershell -File parametertest.ps1

parametertest.ps1 [[-val1] < string>] [[-val2] < Object>] [-val3]

powershell -File parametertest.ps1 sometest 123

val1=sometest val2=123 val3=False

powershell -File parametertest.ps1 -val1 stuff -val2 123 -val3

val1=stuff val2=123 val3=True

powershell -File parametertest.ps1 -unknown parameter

No parameters passed

Found Args -unknown parameter