Passing variable to Linux bash command parameter

Question

I have the following bash script thowring exception: "cut: option requires an argument -- 'f'".

It would cut out the 5th colunm from /etc/passwd, separated by ':' character.

How should I pass the $variable to the option's parameter?

variable=5; cut -d':' -f$variable < /etc/passwd

this should work. make sure you're on bash shell. echo $0 will say. Also same as this variable=5; cut -d: -f$variable /etc/passwd with extra chars eliminated. — karakfa
– karakfa, Commented Mar 5, 2018 at 21:17

glenn jackman · Accepted Answer · 2018-03-05 21:36:59Z

1

You get this when $variable is empty:

$ variable=
$ cut -d: -f$variable </etc/passwd
cut: option requires an argument -- 'f'
Try 'cut --help' for more information.

2 suggestions:

answered Mar 5, 2018 at 21:36

glenn jackman

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Comments

Diego Torres Milano · Accepted Answer · 2018-03-05 21:57:18Z

0

To avoid that, use

set -o nounset

and then, if variable is unset you'll receive

bash: variable: unbound variable

answered Mar 5, 2018 at 21:57

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