Question

I have up to 4 files based on this structure (note the prefixes are dates)

• 0830filename.txt
• 0907filename.txt
• 0914filename.txt
• 0921filename.txt

I want to open the the most recent one (0921filename.txt). how can i do this in a batch file?

Thanks.

Answer 1

This method uses the actual file modification date, to figure out which one is the latest file:

@echo off for /F %%i in ('dir /B /O:-D *.txt') do (     call :open '%%i'     exit /B 0 ) :open     start 'dummy' '%~1' exit /B 0 

This method, however, chooses the last file in alphabetic order (or the first one, in reverse-alphabetic order), so if the filenames are consistent – it will work:

@echo off for /F %%i in ('dir /B *.txt^|sort /R') do (     call :open '%%i'     exit /B 0 ) :open     start 'dummy' '%~1' exit /B 0 

You actually have to choose which method is better for you.