Variable set inside while loop is not visible outside

Question

Variable $adsQ is an output of sql. This variable has number of lines in matrix format. When I pass this variable with pipe to while loop, it works fine whereas I loose variable required from while loop. As suggested from other forums, I modified with followings

            varout=''
            while IFS= read -r adrow;
            do
                    <... do something....>
                    varout="$varout $adrow"
            done < <(printf '%s\n' $adsQ)
            echo "output of while $varout"
            echo "AFTER adsVal >> $adsVal"

when I run this, i get error stating

test.sh: line 72: syntax error near unexpected token <' test.sh: line 72: done < <(printf '%s\n' $adsQ)'

You are running the script with a shell that does not support process substitution, not `bash`. — chepner, Jul 24 '16 at 15:35

score 0 · Answer 1 · answered Jul 25 '16 at 09:42

0

You have an extraneous < in your command. You wrote

done < <(...)

Remove the first <.

answered Jul 25 '16 at 09:42

user1934428

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Variable set inside while loop is not visible outside

1 Answers1