Batch script to Get First Filename from a directory

Question

My function needs the first filename from a particular directory to process some tests using the first file, on completing tests delete first file from directory

I tried as below

FOR /R \\<My Folder> %F in (*.zip*) do echo ~nF  >%test_list%
set /p firstline=%test_list%
echo %firstline% 


FOR /F "tokens=*" %%A IN ('dir %test_firmware_dir% /b/s ^| find /c ".zip"') DO SET 
 count=%%A
  echo %count%
 :test_execution
  if %count% NEQ 0 (
  echo ON
  dir /b %test_firmware_dir%\*.zip >%test_firmware_list%
 set /p firstline=<%test_firmware_list%
 set /a filename=%firstline:~0,-4%
  Echo %filename%
  Echo *********************************************************
 Echo "Now running batch queue tests with %filename%"
  Echo ********************************************************

it is showing last element any procedure to get first element ??

Answer 1

An alternative is to use goto (breaks FOR loops) after you have your first file:

FOR %%F IN (%test_firmware_dir%\*.zip) DO (
 set filename=%%F
 goto tests
)
:tests
echo "%filename%"

And it could run (a little bit) faster in some cases as it doesn't have to go through the whole directory.

Answer 2

Assuming you mean the 1st file when sorted alphabetically; then given that you traverse the list overwriting the last value you will as you say capture the last file, instead control the sort order;

for /f "delims=" %%F in ('dir %test_firmware_dir%\*.zip /b /o-n') do set file=%%F
echo 1st alpha file is %file%