Search code examples
batch-filedos

MSDOS 6.22 How to get batch directory?

I have an old 386 computer (without windows) which has MSDOS 6.22. So I cannot use any solution that built on cmd.exe (part of windows)

I want to pass current bat script path and name to another program within this bat code.

I try to use %CD% but it looks like works only with cmd.exe I try to use %0 argument, but it hold only the name of bat instead of name with full path

@echo off
set myPath=%cd%
myprogram.exe %myPath%\%0

It doesn't work. The passed parameter is \mybat.bat when I started the bat program from it's directory with full name. After the variables resolved, I want to something like this:

@echo off
myprogram C:\BATCH\MYBAT.BAT

Where the bat placed in c:\batch directory, and it name is mybat.bat

How can I do this?

Solution

  • The solution of @Stephan is good, but requires a predefined helper file.

    But you can also build a solution which doesn't need any predefined files.

    @echo off

>temp1.bat echo @PROMPT SET _CD=$P
>temp2.bat command /c temp1.bat
call temp2.bat
del temp1.bat
del temp2.bat

echo currentDir=%_CD%

    This can be used to get the current directory $P, time $T or date $D, as these values are supported by the PROMPT command.