mypath and myfilename

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    by antonela at 2013-03-20 06:06:30

    I have a file c:\temp\myfilename.ps1
    How can I put the path and the name of file in 2 variables ?
    a= "c:\temp"


    by Typeo at 2013-03-20 06:26:19

    You basically had it. You just need to add a $ at the beginning.

    $Path = 'C:\temp'
    $FName = 'Myfilename.ps1'

    by antonela at 2013-03-21 01:49:13

    sorry, I don't want to set the 2 variables.
    I want something that returns the value of file name and the path.

    I want to launch this command:
    powershell.exe c:\temp\myfilename.ps1
    and I want to write a file log c:\temp\myfilename-(

    Is it a way to do this?To put in a variable the name of file I have as an argument?

    by jonhtyler at 2013-03-21 03:56:32

    $filename = split-path -path c:\temp\myfilename.ps1 -leaf
    $newFileName = "$($filename.substring(0, $filename.LastIndexOf('.')))-`($(get-date -format yy.MM.dd)`).log"

    by antonela at 2013-03-21 04:39:46

    Thanks, but I don't want to write c:\temp\myfilename.ps1 in any way in the split-path
    I want to have $filename without writing the name of file, but using a function that returns the name of my file.

    It's important for me because I want to create more cmd:
    powershell myfilename.ps1
    powershell myfilename2.ps1
    powershell myfilename3.ps1

    and have automatically myfilename-( , myfilename2-( , myfilename3-( and so on....
    I thought that if I was able to put in a variable the name of my file, I would came out...

    In every file .ps1 I have an sqlcmd -s .....>>$filelog

    by jonhtyler at 2013-03-21 04:45:13

    So replace "c:\temp\myfilename.ps1" in the split-path cmdlet with a variable of your choice that has the fully-qualified path that you want to process.

    by poshoholic at 2013-03-21 04:57:00

    Sounds like you are trying to get the path of the file from within the script file itself. Here's an example showing how to do that:

    Assume you have a file with the following contents:
    $path = $MyInvocation.MyCommand.Path
    $logFileName = $(Split-Path -Path $Path -Leaf) -replace '\.ps1$',"-$(Get-Date -Format 'yy.MM.dd').log"
    Write-Host "Now we can create a log with a filename of $logFileName"

    Whenever you invoke this, whether dot-sourced, called with the call operator, or invoked using the full path, you'll see that no matter where the file is or what it is named, you'll get a log file using the same name of the file in the format you requested.

    by antonela at 2013-03-21 07:24:45

    Thank you very much
    $path = $MyInvocation.MyCommand.Path
    is what I needed.
    Thanks again!

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