使用批处理文件列出csv上的文件名和文件夹路径

Question

I have a batch file that lists the full file directories (including filename at the end) onto a csv file. I need it to produce this, but also just the filename in a separate column. I would like it so that the format is Filename in first column (including extension) and full file directory in second column. The batch file I currently have is:

dir C:\Users\Administrator\Desktop\test\Files\*.tif /b /s >>
C:\Users\Administrator\Desktop\test\Output.csv

EDIT: I forgot to mention the 'Files' folder contains many subfolders so I need it to process all files from all these subfolders.

Thanks in advance.

Answer 1

You just need to do this:

@echo off
setlocal

set "in=C:\Users\Administrator\Desktop\test\Files\*.tif"
set "out=C:\Users\Administrator\Desktop\test\Output.csv"

if not exist "%out%" type nul>"%out%"
for /f "delims=" %%a in ('dir /b/s %in%') do (
    >>%out% echo "%%~nxa","%%a"
)

Answer 2

Iterate and write ( help for for info on the %~ variable modifiers):

@echo off
cd C:\Users\Administrator\Desktop\test\Files
for %%f in (*.tif) do echo "%%~nxf","%%~dpf" >> Output.csv