UNIX for Dummies Questions & Answers

Hello. A bit of a puzzle here:
I have a 3rd party executable, which requires the following parameters:
parm1 = program_name, parm2=userid/password, parm3=additional flags.
We tried passing password as a variable, but you can do grep, and see what the password actually is
I found a bit different approach, but cannot get it to work
In my script I have something like /mydir/exec prg_name "?" parm3
When I execute my script the "?" is interpreted into prompts: user id and password. How can I pass password without user involvement, and not make it visible via ps?

Your help is greatly appreciated

pipe(), fork(), dup the read part of the pipe on to the child's stdin, exec() in child with the parameters you have, and then write the username and password to the child process's stdin via the pipe.

The username and password WILL be visible in the text of the program, so if you can make it executable but not readable except by root, something like mode 0511? It can be executed by others, but the file can't be read. (I tried this on my openbsd system, and w/ mode 0111, I could run it.)

You may have to figure out what to do with the child's stdout. There are other issues to investigate, possibly, like detaching from the terminal, perhaps, and maybe daemon()'izing it.

HTH

Thank you.
The problem is when doing ps -augxww command
It shows actuall password being used. I have tried with "?" but that requires user input into prompts..
mysrv/ myprogram.ksh
mysrv/ Please Enter UserID:
At this point I type userID <hit enter>
mysrv/ Please Enter Password:
At this point I type password, but it is not visible <hit enter>
At this point my process is executed. In another window I issue ps command from above, and in the processes instead of userid/password I only see ?
So, how do I pass both (login/password) from the file?
I have tried mysrv/ myprogram.ksh < file.txt but only first parm (loginID) is taken and not the password