mypath and myfilename

Welcome Forums General PowerShell Q&A mypath and myfilename

Viewing 0 reply threads
  • Author
    • #5694
      Topics: 1562
      Replies: 0
      Points: 1
      Rank: Member

      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!

Viewing 0 reply threads
  • The topic ‘mypath and myfilename’ is closed to new replies.