return filename

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    by antonela at 2013-03-19 05:05:37

    I have a powershell file (myFile.ps1) with this command (launch a stored procedure):

    sqlcmd -S myserver -d myDB -b -Q "EXEC pr_prova" >"$LogPath\log-$mydate.log"

    1.I want to write a log file with the name myFile-2013.03.18.log (In this moment I have log-2013.03.18.log)
    How could I put in a variable the name of my file .ps1 (myFile)?

    2.Is it possible to create myFile-[2013.03.18].log ? I have an error for the character '['

    $LogPath is a variable that I put in another file Variables.ps1

    In myFile.ps1 I use . .\Variables.ps1

    $LogPath returns "C:\Logs" only f I launch firstly the file Variables.ps1.If not I have an error.....How could I fix this?

    Thank you everybody

    by coderaven at 2013-03-20 12:00:33

    The file can be a dynamic value

    $LogFileName = "$LogPath\log-$(Get-Date -format 'M-d-yyyy').log"

    Just work with the format and Get-Date.

    by MasterOfTheHat at 2013-03-20 13:00:56

    To jump on what Allan has already posted, you would have to escape the brackets if you want them in the file name:
    $LogPath\log-`[$(Get-Date -format 'yyyy-MM-dd')`].log

    by antonela at 2013-03-21 01:38:11

    ok, but I want to put the name 'log' as variable.

    for example
    1. if I have a command c:\windows\system32\powershell.exe c:\my\myfile.ps1
    I want to have a log file with this name: c].log
    2. if I have a command c:\windows\system32\powershell.exe c:\my\myfile2.ps1
    I want to have a log file with this name: c].log

    Is it possible to put in the file log the name of the argument?

    by coderaven at 2013-03-21 11:14:46

    $FillLogpath = "$LogPath-[$(Get-Date -format 'yyyy-MM-dd')]"

    if just running powershell.exe and calling the file, you can pass the values in as normal and inside the script use $args[0] or $args[1] to get the first and second argument respectively. You could also have a param() block as the first line of your file that will receive the arguments.

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