find path of latest file using batch command - batch-file

Question

I have files named file.txt in multiple subfolders in my folder. I want to get the path of the latest file.txt
FOR /r %%i IN ('DIR file.txt /B /O:-D') DO SET a=%%i
echo Most recent subfolder: %a%
gives latest created folder having file.txt whereas I want the folder which has latest file.txt

Answer 1

#echo off
setlocal enableextensions disabledelayedexpansion
for /f "tokens=1,2,*" %%a in ('
robocopy . . file.txt /l /nocopy /is /s /nc /ns /ts /ndl /njh /njs
^| sort /r 2^> nul
^| cmd /v /q /c "set /p .=&echo(^!.^!"
') do (
echo File found : %%c
echo File timestamp (UTC^) : %%a %%b
echo Folder of file : %%~dpc
)
This will use the robocopy command to enumerate all the file.txt files under the current active directory, without copying anything but generating a list of all matching files with a yyyy/mm/dd hh:nn:ss utc timestamp. Then the list is sorted on the timestamp in descending order and only the first line in the output readed and echoed to be processed by the for /f tokenizer and retrieve the date, time and file with full path.
If only the folder is required and the list of files is not very large, a simplified version could be
#echo off
setlocal enableextensions disabledelayedexpansion
for /f "tokens=1,2,*" %%a in ('
robocopy . . file.txt /l /nocopy /is /s /nc /ns /ts /ndl /njh /njs
^| sort /r
') do set "lastFolder=%%~dpc" & goto :done
:done
echo Last folder : %lastFolder%
Almost the same, but instead of including a filter in the list generation to only retrieve the first line (required if the list of files is very large), here the for /f will retrieve the full list but after the first element is processed we jump out of the loop to the indicated label.