Hi, I have a directory set up that stores *.CSV files. I need to read the file name of the file, create a directory with the same file name, move the *.csv file into the new directory and rename the file to the date and time.CSV

rename the file to the date and time.CSV Do you want the current date/time or the file creation date/time?

Hi, Yes I do. I have found this that nearly fits the bill for /f %%F in ('dir/b/a-d *.csv') do call :sub1 %%F goto :eof :sub1 set name=%1 md %name:~0,6% move %* %name:~0,6% but only looks for a specific string length of file name rather than *.CSV. Can you help please?

You didn't answer my query on date/time so I've used the file created date and time. Without knowing your date/time format the script is set up for my format which is dd-mm-yyyy hh:mm. You will have to change set file=!file:~6,4!!file:~3,2!!file:~0,2!_!file:~11,2!!file:~-2!.csv to suit your format if it is different. The output files are named in the format yyyymmdd_hhmm. Once you have finished testing change the "Copy" command to "Move". The script is untested. @echo off cls setlocal enabledelayedexpansion for /f "delims=" %%A in ('dir /a-d /b /tc "*.csv"') do ( md "%%~nA" set file=%%~tA set file=!file:~6,4!!file:~3,2!!file:~0,2!_!file:~11,2!!file:~-2!.csv copy "%%A" "%%~nA\!file!"> nul ) Good luck

Sorry about not answering your question properly. Your script works as indicated but I have changed the final date and time format of the CSV file. Many thanks for your time to answer my question.