Bash - assign array into variable as string

Question

I have this code, it prints the correct result, but I can't figure out how to get the echo from the last line into a variable.

# hostname is 'tech-news-blog-324344' . Setting it into array host_name_array
IFS='-' read -r -a host_name_array <<< "$(hostname)" 
#removing the last part of string after last "-"
unset 'host_name_array[${#host_name_array[@]}-1]'
( IFS=$'-'; echo "${host_name_array[*]}" )  
#result is 'tech-news-blog'

How could get the value of the last line into a variable? I've tried the following:

( IFS=$'-'; URL="${host_name_array[*]}" )

But I get the result "tech news blog" with spaces between pieces of array instead of '-'.

Solution 1

When IFS='-' read -r -a host_name_array <<< "$(hostname)" is ran, the array is (tech news blog 324344).

After the final element is removed with unset 'host_name_array[${#host_name_array[@]}-1]', the array is (tech news blog).

So, to get this to echo tech-news-blog, some substitution will have to be done, since echo "${host_name_array[*]}" will yield tech news blog:

With tr: echo "${host_name_array[*]}" | tr ' ' '-'

sed: echo "${host_name_array[*]}" | sed 's/ /-/g'

Solution 2

You could do this by outputting the elements separated by -, and then strip off the last - by parameter expansion:

$ var=$(printf '%s-' "${host_name_array[@]}")

$ echo "$var"
foo-bar-spam-egg-

$ var="${var%-}"

$ echo "$var"
foo-bar-spam-egg

Also you need ${host_name_array[@]} instead of ${host_name_array[*]} to prevent outputting the elements as a whole separated by first character of IFS.

---

What you are trying to do can be achieved by a simple parameter expansion:

${var%-*}

Example:

$ var=$(hostname)

$ echo "${var%-*}"

Solution 3

(...) introduces a subshell. So $URL wouldn't be set after that (still have the value it had before the subshell). You want:

IFS=-
read -r -a host_name_array <<< "$(hostname)" 
unset 'host_name_array[${#host_name_array[@]}-1]'
URL="${host_name_array[*]}"

"${host_name_array[*]}" joins the elements on the arrays on the first character of $IFS just like "$*" does in standard sh.

If the reason why you're using a subshell is because you don't want to modify $IFS globally, you could do that in a function where you give $IFS a local scope:

f() {
  local IFS=-
  ...
}
f

Or use command substitution that also creates a subshell but allows passing data to the parent shell:

URL=$(IFS=-; printf '%s\n' "${host_name_array[*]}")

Here though I'd use standard sh expansion to remove that trailing component:

URL=$(uname -n) # uname -n is the standard equivalent of hostname
URL=${URL%-*}

It has several advantages over the above:

• it works even if the text contains newline characters (very unlikely for host names here though);
• if the text doesn't contain -, it doesn't remove anything;
• it is more efficient. read <<< $(hostname) means running hostname, read its output through a pipe, storing that in a temp file and have read read the first line of that;
• it doesn't clobber the variable namespace with those temporary variables
• it is shorter;
• it is portable. No need to install bash to run that code. The system's sh will suffice.

In any case, remember to quote your variables when using them in list contexts:

printf '%s\n' "$URL"

When doing:

echo $URL

You're invoking the split+glob operator. So, if $IFS still contains -, that $URL would be split on -, and echo would output those words separated by spaces.

Comments

[BenB]

Both the echo works, but if I try to set that result into a variable and reuse it I get from the variable the string with spaces

[Kenneth B. Jensen]

var=$(echo "${host_name_array[*]}" | tr ' ' '-'), but I think @heemayl's solution is better, since it is much less bulky and awkward.

[BenB]

When I echo the $var I get the string with spaces for some reason

[BenB]

About his answer, I have the same issue there when I echo the $var I get the string with spaces

[Kenneth B. Jensen]

Is IFS still set to '-'? If so, that may be the cause; reset it with unset IFS.

[F. Hauri]

Wow, this is strong! Why did you not use printf -v var "%s-" ... syntax?

[F. Hauri]

@heemayl Using var=$(printf... implie one fork to subshell, where printf -v ... let you use builtin printf and set var without forks! This is a lot quicker.

[heemayl]

@F.Hauri I was serious there :) You are absolutely right :)