How to pass arguments to /bin/bash? [closed]

问题

In my terminal, I do $ myprogram myfile.ext.

I want to be able to do the same via the command /bin/bash, but it is not working.

I tried

$ /bin/bash -c -l myprogram myfile.ext

but it is not working. It seems that my program is launched but without my file.ext as an option (as an input file)

Does anyone has a clue on how to do that?

Why I ask this question

I want to launch myprogram via NSTask in an a program I am writing for OSX, with the cocoa framework. If I simply ask NSTask to launch my program, it seems that some variable environment are missing. So, what I try is to launch a shell and inside this shell to launch myprogram.

回答1:

Drop the -c altogether. From the manpage:

bash [options] [file]
…
Bash  is  an  sh-compatible command language interpreter that executes commands read from the standard input or from a file.

Thus, you can execute your program via

bash myprogram myfile.ext

and myfile.ext will be the first positional parameter.

bash -l myprogram myfile.ext

will work as well, (but whether or not bash is invoked as a login shell seems tangential to this question.)

回答2:

-c string 
    If  the  -c  option  is  present, then commands are read from
    string.  If there are arguments after the  string,  they  are
    assigned to the positional parameters, starting with $0.

You need to quote the command passed in as string so it's treated as a single argument for the -c option, which then gets executed normally with the argument following the command, i.e.

/bin/bash -c -l 'myprogram myfile.ext'

来源：https://stackoverflow.com/questions/23025355/how-to-pass-arguments-to-bin-bash

标签

bash

cocoa

unix

terminal

nstask