bash while read loop arguments from variable issue

Question

I have a bash script with following variable:

operators_list=$'andrii,bogdan,eios,fre,kuz,pvm,sebastian,tester,tester2,vincent,ykosogon'

while IFS=, read -r tech_login; do


    echo "... $tech_login ..."

done <<< "$operators_list"

I need to read arguments from variable and work with them in loop. But it returns echo only one time with all items:

+ IFS=,
+ read -r tech_login
+ echo '... andrii,bogdan,eios,fre,kuz,pvm,sebastian,tester,tester2,vincent,ykosogon ...'
... andrii,bogdan,eios,fre,kuz,pvm,sebastian,tester,tester2,vincent,ykosogon ...
+ IFS=,
+ read -r tech_login

What am I doing wrong? How to rework script, so it will work only with one item per time?

Answer 1

operators_list=$'andrii,bogdan,eios,fre,kuz,pvm,sebastian,tester,tester2,vincent,ykosogon'

So you have strings separated by ,. You can do that multiple ways:

1. using bash arrays:

IFS=, read -a operators <<<$operators_list
for op in "${operators[@]}"; do
    echo "$op"
done

2. Using a while loop, like you wanted:

while IFS= read -d, -r op; do
    echo "$op"
done <<<$operators_list

3. Using xargs, because why not:

<<<$operators_list xargs -d, -n1 echo

The thing with IFS and read delimeter is: read reads until delimeter specified with -d. Then after read has read a full string (usually whole line, as default delimeter is newline), then the string is splitted into parts using IFS as delimeter. So you can:

while IFS=: read -d, -r op1 op2; do
    echo "$op1" "$op2"
done <<<"op11:op12,op12:op22"