How to set the working directory of a command in a Windows batch file?

Question

Let's say I have these commands:

Prog1.exe D:\SomeDir\Prog2.exe Prog3.exe 

Now, say for the second line, I would like the working directory to be D:\SomeDir, but in Prog1.exe and Prog3.exe I want the default working directory (normally, where my .bat file is). If I try this

Prog1.exe cd D:\SomeDir D:\SomeDir\Prog2.exe Prog3.exe 

Apparently Prog3 will be executed in SomeDir, which is not what I want.

Answer 1

You could use the pushd/popd commands (help with pushd /?)

Prog1.exe Pushd D:\SomeDir Prog2.exe popd Prog3.exe 

Answer 2

You could use the cd command (help with cd /?) with the %~dp0, batch file path, variable.

Prog1.exe cd D:\SomeDir Prog2.exe cd %~dp0 Prog3.exe 

For a complete list of %~ modifiers see call /? or for /? help.

However, I only add this as to provide a more complete answer on Stack Overflow. I would RECOMMEND using jeb's solution above.