在另一个* .sh文件的上下文中执行代码

Question

I am a newbie in these matters and feel kind of lost.

My System is Windows 7.

I have an external initialization file init.sh :

# Some code...

exec "${BASH}" --login -i

Now I am writing my own file myFile.sh which starts by calling init.sh and which should execute further code in the bash context of init.sh . I tried:

exec "init.sh"

# Further code here

Which is not working. I would be grateful for any help.

Answer 1

exec instantly replaces a process with whatever was exec -ed. You can't run code after an exec , because the code that was running before the exec is gone , replaced by the new process. Normally, replacing exec "init.sh" with . "init.sh" . "init.sh" (or source "init.sh" ) would allow you to run the contents of init.sh in the current shell without replacing the shell, but because init.sh itself includes an exec , that won't work.

Given that init.sh is running in interactive mode, it doesn't make sense to execute further code after that in any event; a shebang ( #!init.sh ) line could be used to make init.sh the interpreter for the remaining code, but since it's interactive, that doesn't make sense.

It sounds an awful lot like you're trying to reinvent .bashrc / .bash_profile / .bash_login files here; you might want to look at those instead of trying to simulate them, badly.

Answer 2

You can use the source command:

source "init.sh"

This will run the commands in the referenced script ("init.sh") as if its contents were in this script.

. "init.sh"

is a short-hand.