CODEWITHSUNDEEP
.netpowershelldirectory
pencilCake
pencilCake

Reputation: 53263

How can I get all the file names in a directory as a comma separated string in Powershell?

I want to get all the names in a directory as a comma separated string so that I can pass them as a parameter to a cmdlet.

How can I get all the file names in a directory as a comma separated string in Powershell?

Upvotes: 6

Views: 6675

Answers (4)

charlie arehart
charlie arehart

Reputation: 6884

For those who may be more comfortable with the classic Windows dir, here's a variant of Eddie's answer (which used the *nix-like ls, which of course ps also supports):

(dir -name) -join ','

This would produce the exact same result as his ls example. (To be clear, this would work ONLY in powershell--as requested in the question here--but not in the Windows Command line.)

And if you wanted to remove folder names from the results (in either his or my example), add the -file switch: (dir -file -name) -join ',' or (ls -file).name -join ','.

And of course you could follow the dir (or ls) with a pattern of files to be found, or a folder in which to look. These examples are simply getting ALL files in the current folder. To find all filename of exe's starting j, this would work:

(dir j*.exe -file -name) -join ','

Or to name a path as well, note that if the folder has spaces you must quote the path, as in:

(dir "C:\Program Files\Java\jdk-17\bin\j*.exe" -file -name) -join ','

FWIW, this dir in powershell is an alias of Get-ChildItem, and using help dir will show its many other args/switches.

Upvotes: 0

Eddie Kumar
Eddie Kumar

Reputation: 1490

Quickest/shortest way:

(ls).name -join ','

For specific folder-path:

(ls path\to\folder).name -join ','

Upvotes: 4

EBGreen
EBGreen

Reputation: 37750

This should work:

(ls C:\PATH\TO\FOLDER | select -expandproperty name) -join ','

If there are subfolders in there that you want to avoid:

(ls C:\PATH\TO\FOLDER | ?{$_.PSIsContainer -eq $false} | select -expandproperty name) -join ','

Upvotes: 2

CB.
CB.

Reputation: 60938

one way is:

 (dir  | % { $_.basename }) -join ','

Upvotes: 9

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