Copy specific files from a folder with powershell

Question

I need to copy lots of specific pdf files with specific names. All the files are insame folder, I have all the file names in csv. I made a lazy "script" which is like this (obv. not all the files in this post):

Copy-Item -Path 391248* -Destination <destination>
Copy-Item -Path 241044* -Destination <destination>
Copy-Item -Path 532350* -Destination <destination>

Does anyone have more clean solution for this?

Does this answer your question? Recursively copy a set of files from one directory to another in PowerShell — Jacques Koekemoer
– Jacques Koekemoer, Commented Dec 18, 2019 at 9:30

f6a4 · Accepted Answer · 2019-12-18 09:45:43Z

0

$fileList  = 'C:\filelist.csv'

Get-Content -Path $fileList | ForEach-Object {

    Copy-Item -Path $_ -Destination 'destination'
}

answered Dec 18, 2019 at 9:45

f6a4

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2 Comments

Arun Vinoth PrecogTechnologies Over a year ago

Please dont post code only answer, try to add some context of how this will solve the problem in question.

f6a4 Over a year ago

This is a simple one liner only

avarice · Accepted Answer · 2019-12-18 10:06:07Z

0

Another one liner solution is:

cat "The path of csv file.csv" | % {copy -path $_ -destination "The destination"}

answered Dec 18, 2019 at 10:06

avarice

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Comments

Vish · Accepted Answer · 2019-12-18 14:05:29Z

My CSV looks like this:

Filename
1
2
3
4
5
6
7
8
9

This csv does not contain the file extension, the below code adds the file extension '.pdf'. You can modify it to contain anything you require.

$csv_location = 'path to your csv'

# 'FileName' is the name of the column inside the csv
$file_names   = ( Import-Csv $csv_location ).FileName
$destination  = 'path of files to be copied into'
$origin       = 'path of files to be copied from'

$file_names | ForEach-Object {

#   This line adds the extension '.pdf' to the filename and file path obtained from $file_names and $origin, respectively.
    $full_path = $origin + $_ + '.pdf'

#   Copy from origin to ( destination + file name + file extension )
    Copy-Item -Path ( $full_path ) -Destination $destination -Force
}

If your csv contains the file extension,

$full_path = $origin + $_

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