Missing '=' operator after key in hash literal

Thanks for your reply.  If I include the source and destination inside an @array, the script works fine. It did not work when the source and destination file is in a CSV file and I import it.  Here is how I modified the script using an array:

FilesAndFolders = @{ "C:\Data_Files\fileone.$TodayDate" = "E:\DEV\_Datafiles\fileone_Extracted"
"C:\Data_Files\filetwo.$TodayDate" = "E:\DEV\_Datafiles\filetwo_Extracted"
"C:\Data_Files\filethree.$TodayDate" = "E:\DEV\_Datafiles\filethree_Extracted"
"C:\Data_Files\filefour.$TodayDate" = "E:\DEV\_Datafiles\filefour_Extracted"
"C:\Data_Files\filefive.$TodayDate" = "E:\DEV\_Datafiles\filefive_Extracted"}

foreach ($KeyAndValue in $ZipFilesAndFolders.GetEnumerator())
{
$ZipFilePath = $KeyAndValue.Key
$DestinationPath = $KeyAndValue.Value

echo "-------------------------------------------"
echo "start extracting"
Get-ChildItem $ZipFilePath | % {& "E:\7-Zip\7za.exe" "x" $_.fullname "-o$DestinationPath"}
echo "Finish extracting $Dept_name"
echo "--------------------------------------------"
}

With this approach the script is working but I dont want to to that. I want to keep the source and destination into a CSV file because I only gave 4 files as example but in reality i have 215 files to extracted. That is why I want to hold it in CSV file and import it.  Could you please help?

One more thing : How do you collect information for you .csv file ?

My approach will be:

Get-ChildItem -Path \\path\to\*.zip -file | select-object -property FullName | Export-Csv -Path \\path\to\ZipFileFolders.csv -noTypeInformation (Optional : -delimiter -encoding)

With this, you'll have a correct .csv file, prepare for use. Nota if you use optional params with Export-csv use it again with Import-csv.

Bye. It's late, it's Hot (30°c in my bedroom), but it's time to go to bed. I'm get up like the sun, early in the morning (or late in the night depending of how you see the bottle) :-)

Oliv

Thank you very much for your input. It make sense to me now. The only things i want to know is that I have Powershell 4.0 and don't have expand-archive cmdlet.  I use 7-Zip instead. How am I going to change this line below to 7-Zip ? That is the only part i want to know. Thank you in advance.

Expand-Archive -Path $file.SourceFile -DestinationPath  $file.DestinationFolder

I managed to make the script working. It only work if I hard-coded the date (DT190829). If I have the variable $TodayDate the script did not understand and says file could not find. I want to used $TodayDate because I dont want to change the date manually. How to do that?  Below is the script that is working.

##Here is the external .CSV file as you suggested.
SourceFile, DestinationFolder
C:\Data_Files\fileone.$TodayDate, E:\DEV\_Datafiles\fileone_Extracted
C:\Data_Files\filetwo.$TodayDate, E:\DEV\_Datafiles\filetwo_Extracted
C:\Data_Files\filethree.$TodayDate, E:\DEV\_Datafiles\filethree_Extracted

##Unzip function
Add-Type -AssemblyName System.IO.Compression.FileSystem
function Unzip
{
  param([string]$zipfile, [string]$outpath)
  [System.IO.Compression.ZipFile]::ExtractToDirectory($zipfile, $outpath)
}

$TodayDate = (Get-Date).ToString('DTyyMMdd')

$ZipFiles  = Import-csv "Y:\PowershellScript\Development\ZipFileFolders.csv"
 
   foreach ($file in $ZipFiles)
   {
       
       ## took the first 3 characters of the extracted files
        $Dept_name = $file.DestinationFolder.substring(44,3)
        echo " departname is: $Dept_name"

        echo "-------------------------------------------"
        echo "start extracting $Dept_name"
        Unzip $file.SourceFile   $file.DestinationFolder  
       echo "Finish extracting $Dept_name"
       echo "--------------------------------------------"
  }