How can I get the current PowerShell executing file?

Note: PowerShell 1.0
I'd like to get the current executing PowerShell file name. That is, if I start my session like this:

powershell.exe .\myfile.ps1

I'd like to get the string ".\myfile.ps1" (or something like that). EDIT: "myfile.ps1" is preferable.
Any ideas?

I've tried to summarize the various answers here, updated for PowerShell 5:

• If you're only using PowerShell 3 or higher, use $PSCommandPath

• If want compatibility with older versions, insert the shim:

  if ($PSCommandPath -eq $null) { function GetPSCommandPath() { return $MyInvocation.PSCommandPath; } $PSCommandPath = GetPSCommandPath; }

  This adds $PSCommandPath if it doesn't already exist.

  The shim code can be executed anywhere (top-level or inside a function), though $PSCommandPath variable is subject to normal scoping rules (eg, if you put the shim in a function, the variable is scoped to that function only).

Details

There's 4 different methods used in various answers, so I wrote this script to demonstrate each (plus $PSCommandPath):

function PSCommandPath() { return $PSCommandPath; }
function ScriptName() { return $MyInvocation.ScriptName; }
function MyCommandName() { return $MyInvocation.MyCommand.Name; }
function MyCommandDefinition() {
    # Begin of MyCommandDefinition()
    # Note: ouput of this script shows the contents of this function, not the execution result
    return $MyInvocation.MyCommand.Definition;
    # End of MyCommandDefinition()
}
function MyInvocationPSCommandPath() { return $MyInvocation.PSCommandPath; }

Write-Host "";
Write-Host "PSVersion: $($PSVersionTable.PSVersion)";
Write-Host "";
Write-Host "`$PSCommandPath:";
Write-Host " *   Direct: $PSCommandPath";
Write-Host " * Function: $(PSCommandPath)";
Write-Host "";
Write-Host "`$MyInvocation.ScriptName:";
Write-Host " *   Direct: $($MyInvocation.ScriptName)";
Write-Host " * Function: $(ScriptName)";
Write-Host "";
Write-Host "`$MyInvocation.MyCommand.Name:";
Write-Host " *   Direct: $($MyInvocation.MyCommand.Name)";
Write-Host " * Function: $(MyCommandName)";
Write-Host "";
Write-Host "`$MyInvocation.MyCommand.Definition:";
Write-Host " *   Direct: $($MyInvocation.MyCommand.Definition)";
Write-Host " * Function: $(MyCommandDefinition)";
Write-Host "";
Write-Host "`$MyInvocation.PSCommandPath:";
Write-Host " *   Direct: $($MyInvocation.PSCommandPath)";
Write-Host " * Function: $(MyInvocationPSCommandPath)";
Write-Host "";

Output:

PS C:\> .\Test\test.ps1

PSVersion: 5.1.19035.1

$PSCommandPath:
 *   Direct: C:\Test\test.ps1
 * Function: C:\Test\test.ps1

$MyInvocation.ScriptName:
 *   Direct:
 * Function: C:\Test\test.ps1

$MyInvocation.MyCommand.Name:
 *   Direct: test.ps1
 * Function: MyCommandName

$MyInvocation.MyCommand.Definition:
 *   Direct: C:\Test\test.ps1
 * Function:
    # Begin of MyCommandDefinition()
    # Note this is the contents of the MyCommandDefinition() function, not the execution results
    return $MyInvocation.MyCommand.Definition;
    # End of MyCommandDefinition()


$MyInvocation.PSCommandPath:
 *   Direct:
 * Function: C:\Test\test.ps1

Notes:

• Executed from C:\, but actual script is C:\Test\test.ps1.
• No method tells you the passed invocation path (.\Test\test.ps1)
• $PSCommandPath is the only reliable way, but was introduced in PowerShell 3
• For versions prior to 3, no single method works both inside and outside of a function

While the current Answer is right in most cases, there are certain situations that it will not give you the correct answer. If you use inside your script functions then:

$MyInvocation.MyCommand.Name 

Returns the name of the function instead name of the name of the script.

function test {
    $MyInvocation.MyCommand.Name
}

Will give you "test" no matter how your script is named. The right command for getting the script name is always

$MyInvocation.ScriptName

this returns the full path of the script you are executing. If you need just the script filename than this code should help you:

split-path $MyInvocation.PSCommandPath -Leaf

If you only want the filename (not the full path) use this:

$ScriptName = $MyInvocation.MyCommand.Name

Try the following

$path =  $MyInvocation.MyCommand.Definition 

This may not give you the actual path typed in but it will give you a valid path to the file.