Passing arguments in unix in command line

Question

How can I pass arguments to a command in unix? For example, if I have to open a file:

R> vi john/pic/mars/NASA/rover.txt

In the above vi command, I want to replace "mars" with a variable, and pass the variable value in the same line, as in:

R> vi john/pic/$variable/NASA/rover.txt | $varaiable=pluto

Of course this doesn't work. But I hope my question is clear. Can anyone help me with this?

@n.m. I would have thought so, too, but the variable actually only gets defined in the launched process itself, not the current command-line (in Bash 4.2.45(1); probably a feature). Have you actually tried it (without the typo?). — Ingo Karkat
– Ingo Karkat, Commented Apr 8, 2014 at 12:12

Rubens · Accepted Answer · 2014-04-12 23:04:10Z

Simply move the variable definition to the beginning of the command line, as in:

variable=pluto; vi john/pic/$variable/NASA/rover.txt

or even:

variable=pluto && vi john/pic/$variable/NASA/rover.txt

OBS:

notice you can't use $ while defining variables, only while using their values;
piping your vi command to a variable assignment does not make much sense, although you could achieve some clearer parameterization from:
```
function opener() { vi john/pic/$1/NASA/rover.txt } $ opener "pluto" 
```

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