I saw similar questions but nothing is working yet.
The end goal is to get a csv with col1 being file path and col2 being creation date.
Lets assume i have a directory C:/dir1, and inside it I have C:/dir1a and C:/dir1b and dir1a and dir1b both have files in them.
I can do
dir /s /b > dirfiles.csv
Which does give me all the file names, however when I do
dir /S /B /T:C > dirfiles_time.csv
nothing different happens.
How can i create a list of paths to every file (not directory just the files) and the creation time of that file?
Thanks
EDIT I do not care about the actual directory, only the path to the files themselves. Remove /b leaves a lot of information that i would need to remove
Directory of D:\Dir1\dir1a
03/29/2018 11:20 AM <DIR> .
03/29/2018 11:20 AM <DIR> ..
03/29/2018 11:20 AM 5,583,992 16385 yadayada.tif
03/29/2018 11:20 AM 9,560,580 2278 yada.jpg
2 File(s) 15,144,572 bytes
To fit in with your [batch-file]
tag, just run this PowerShell
command from the batch file.
@Powershell -NoP -C "GCI $(GCI \"D:\Dir1\"|?{$_.PSIsContainer}|%%{$_.FullName}) -R|?{!$_.PSIsContainer}|Select FullName,CreationTime|Export-CSV -NoT \"dirfiles_time.csv\""
You can change the initial directory, D:\Dir1
, and csv output file, dirfiles_time.csv
, just take care not to delete their opening and closing, \"
sequence.
If you want only the filenames without their paths then change Select FullName,CreationTime
to Select Name,CreationTime
. Likwise If you'd prefer the last modified times, change Select FullName,CreationTime
to Select FullName,LastWriteTime
.