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.