How to concatenate multiple shell variables into one?

Question

I need to form a single variable from a for loop.

My script:

#! /bin/sh
if [ "$#" -le 1 ]; then
    echo "Illegal number of parameters"
    exit 1
else
    # Form domains variable
    DOMAINS=""
    for i in "${@:2}"; do
        DOMAINS+="$i,"
    done

    echo "${DOMAINS::-1}"
fi

When I execute it:

    sh script.sh command domain1 domain2

I get the following error:

    certbot.sh: line 10: DOMAINS+=mmand,: not found
    certbot.sh: line 10: DOMAINS+=domain1.com,: not found
    certbot.sh: line 10: DOMAINS+=domain2.com,: not found

It seems as I used bash syntax since the following execution works:

    bash script.sh command domain1.com domain2.com

I get:

domain1.com,domain2.com

I need it to work as sh not bash. I can't seem to find a solution.

Answer 1

+= is not valid in a POSIX shell.

Since it is not a valid variable assignment,

DOMAINS+="$i,"

is interpreted as the name of a command, which is obtained by parameter expansion of i . For instance, if i equals 1, the line corresponds to

DOMAINS+=1,

If you had an executable file named DOMAINS+=1, in your PATH, this file would be run.

You have to catenatate variables like this:

FOO=$FOO$BAR$BAZ

You can't avoid repeating the name FOO .

An alternative would be to switch to zsh or bash, where your usage of += would indeed have the desired effect.

Answer 2

Just:

IFS=,
echo "$*"

Or you seem to want from a second argument. Then like:

( shift; IFS=,; echo "$*" )

Answer 3

Did you try changing DOMAINS+="$i," into DOMAINS="${DOMAINS}${i}," relying on variable substitution?