Convert multiline string to array

Question

I have this script:

nmapout=`sudo nmap -sP 10.0.0.0/24`
names=`echo "$nmapout" | grep "MAC" | grep -o \'(.\\+)\'`
echo "$names"
         


        

Answer 1

Bash also has a readarray builtin command, easily searchable in the man page. It uses newline (\n) as the default delimiter, and MAPFILE as the default array, so one can do just like so:

    names="Netgear
    Hon Hai Precision Ind. Co.
    Apple"

    readarray -t <<<$names

    printf "0: ${MAPFILE[0]}\n1: ${MAPFILE[1]}\n2: ${MAPFILE[2]}\n"

The -t option removes the delimiter ('\n'), so that it can be explicitly added in printf. The output is:

    0: Netgear
    1: Hon Hai Precision Ind. Co.
    2: Apple

Answer 2

As others said, IFS will help you.IFS=$'\n' read -ra array <<< "$names" if your variable has string with spaces, put it between double quotes. Now you can easily take all values in a array by ${array[@]}

Answer 3

Let me contribute to Sanket Parmar's answer. If you can extract string splitting and processing into a separate function, there is no need to save and restore $IFS — use local instead:

#!/bin/bash

function print_with_line_numbers {
    local IFS=$'\n'
    local lines=($1)
    local i
    for (( i=0; i<${#lines[@]}; i++ )) ; do
        echo "$i: ${lines[$i]}"
    done
}

names="Netgear
Hon Hai Precision Ind. Co.
Apple"

print_with_line_numbers "$names"

See also:

• Setting IFS for a single statement

Answer 4

Set IFS. Shell uses IFS variable to determine what the field separators are. By default IFS is set to the space character. Change it to newline.

#!/bin/bash
names="Netgear
Hon Hai Precision Ind. Co.
Apple"

SAVEIFS=$IFS   # Save current IFS
IFS=$'\n'      # Change IFS to new line
names=($names) # split to array $names
IFS=$SAVEIFS   # Restore IFS

for (( i=0; i<${#names[@]}; i++ ))
do
    echo "$i: ${names[$i]}"
done

Output

0: Netgear
1: Hon Hai Precision Ind. Co.
2: Apple