Using a BASH script variable into an executed command

Question

Can someone please help with this because I can't seem to find a solution. I have the following script that works fine:

#!/bin/bash

#Checks the number of lines in the userdomains file

NUM=`awk 'END {print NR}' /etc/userdomains.hristian`;

echo $NUM


#Prints out a particular line from the file (should work with $NUM eventually)

USER=`sed -n 4p /etc/userdomains.hristian`

echo $USER


#Edits the output so that only the username is left

USER2=`echo $USER | awk '{print $NF}'`

echo $USER2

However, when I substitute the 4 on line 12 with the variable $NUM, like this, it doesn't work:

USER=`sed -n $NUMp /etc/userdomains.hristian`

I tried a number of different combinations of quotes and ${}, however none of them seem to work because I'm a BASH newbie. Help please :)

You didn't try all the combinations: USER=`sed -n ${NUM}p /etc/userdomains.hristian`. As a side note: don't use upper case variable names in Bash! here you're doing it wrong: USER is very likely a variable set and used by your system; you might have a clash sooner or later! As a bonus note: don't use backticks, use $(...) instead: USER=$(sed -n ${NUM}p /etc/userdomains.hristian). — gniourf_gniourf
– gniourf_gniourf, Commented Feb 7, 2015 at 14:27
Are you trying to print the last line of your file? if so, tail -n1 /etc/userdomains.hristian will do the job more efficiently. — gniourf_gniourf
– gniourf_gniourf, Commented Feb 7, 2015 at 14:34
The last field of the last line, apparently. The shortest way to do that is probably awk 'END { print $NF }' /etc/userdomains.hristian, although using tail -n1 /etc/userdomains.hristian | awk '{ print $NF }' will be more efficient for large files. — Wintermute
– Wintermute, Commented Feb 7, 2015 at 14:37

Tom Fenech · Accepted Answer · 2015-02-07 14:34:44Z

2

I'm not sure exactly what you've already tried but this works for me:

$ cat out
line 1
line 2
line 3
line 4
line 5
$ num=4
$ a=`sed -n ${num}p out`
$ echo "$a"
line 4

To be clear, the issue here is that you need to separate the expansion of $num from the p in the sed command. That's what the curly braces do. Note that I'm using lowercase variable names. Uppercase ones should be be reserved for use by the shell. I would also recommend using the more modern $() syntax for command substitution:

a=$(sed -n "${num}p" out)

The double quotes around the sed command aren't necessary but they don't do any harm. In general, it's a good idea to use them around expansions.

Presumably the script in your question is a learning exercise, which is why you have done all of the steps separately. For the record, you could do the whole thing in one go like this:

awk 'END { print $NF }' /etc/userdomains.hristian

In the END block, the values from the last line in the file can still be accessed, so you can print the last field directly.

edited Feb 7, 2015 at 14:34

answered Feb 7, 2015 at 14:27

Tom Fenech

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1 Comment

Neekoy Over a year ago

Oh my, it works. Thank you very much for the information and the recommendation. Note taken.

QuiteClose · Accepted Answer · 2015-02-07 14:28:38Z

1

Your trying to evaluate the variable $NUMp rather than $NUM. Try this instead:

USER=`sed -n ${NUM}p /etc/userdomains.hristian`

answered Feb 7, 2015 at 14:28

QuiteClose

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