Bash - How to put variable in command?

Question

Which is the righy way to put a variable inside a command in bash?

I'm trying with:

PORT=80
`nc -zv -w30 127.0.0.1 $PORT >> /dev/null`

but it doesn'work.

maybe don't send the output to /dev/null?

Kevin
– Kevin

2011-10-31 22:00:45 +00:00
Commented Oct 31, 2011 at 22:00 — Kevin
– Kevin, Commented Oct 31, 2011 at 22:00

mwp · Accepted Answer · 2011-10-31 22:03:11Z

2

You don't need to use backticks if you're not capturing the output of the command. Just run the command.
If you're putting the output to devnull, you don't need to append (>>), just write (>).
That should work. If it's not working, something else is wrong.
```
PORT=80
nc -zv -w30 127.0.0.1 $PORT > /dev/null
```

answered Oct 31, 2011 at 22:03

mwp

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2 Comments

Mark Over a year ago

It works, but it still shows the output (> /dev/null not working)

A.H. Over a year ago

Then either remove the -v option or redirect stderr like this: >/dev/null 2>&1 .

Kevin · Accepted Answer · 2011-10-31 22:03:30Z

0

I assume you mean you want the output of the command stored in a variable. If so then you should first of all assign the command to a variable, and secondly don't send the output to /dev/null.

x=`nc -zv -w30 127.0.0.1 $PORT`

OR alternate syntax:

x=$(nc -zv -w30 127.0.0.1 $PORT)

answered Oct 31, 2011 at 22:03

Kevin

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