How do I split a string on a delimiter in Bash?

Question

I have this string stored in a variable:

IN="[email protected];[email protected]"

Now I would like to split the strings by ; delimiter so that I have:

ADDR1="[email protected]"
ADDR2="[email protected]"

I don't necessarily need the ADDR1 and ADDR2 variables. If they are elements of an array that's even better.

---

After suggestions from the answers below, I ended up with the following which is what I was after:

#!/usr/bin/env bash

IN="[email protected];[email protected]"

mails=$(echo $IN | tr ";" "\n")

for addr in $mails
do
    echo "> [$addr]"
done

Output:

> [[email protected]]
> [[email protected]]

There was a solution involving setting Internal_field_separator (IFS) to ;. I am not sure what happened with that answer, how do you reset IFS back to default?

RE: IFS solution, I tried this and it works, I keep the old IFS and then restore it:

IN="[email protected];[email protected]"

OIFS=$IFS
IFS=';'
mails2=$IN
for x in $mails2
do
    echo "> [$x]"
done

IFS=$OIFS

BTW, when I tried

mails2=($IN)

I only got the first string when printing it in loop, without brackets around $IN it works.

Answer 1

You can set the internal field separator (IFS) variable, and then let it parse into an array. When this happens in a command, then the assignment to IFS only takes place to that single command's environment (to read ). It then parses the input according to the IFS variable value into an array, which we can then iterate over.

IFS=';' read -ra ADDR <<< "$IN"
for i in "${ADDR[@]}"; do
    # process "$i"
done

It will parse one line of items separated by ;, pushing it into an array. Stuff for processing whole of $IN, each time one line of input separated by ;:

 while IFS=';' read -ra ADDR; do
      for i in "${ADDR[@]}"; do
          # process "$i"
      done
 done <<< "$IN"