bash command arguments

I have a bash script that, for reasons I won't discuss, cannot be made executable. However, I need to pass arguments to that script.

I have开发者_如何学Go tried this:

bash MyBashScript.sh MyArgumentOne

But the argument MyArgumentOne isn't passed to the script. I know there must be a way to do this, can anyone help?

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Your given command should work. Try to debug with calling trough

strace -o all_system_calls.txt -f -ff bash MyBashScript.sh MyArgumentOne

one of the all_system_calls.txt.<pid> files created should contain something like

execve("/bin/bash", ["bash", "MyBashScript.sh", "MyArgumentOne"], [/* 71 vars */]) = 0

If so you know for sure that the argument is passed into your script.