BASH - Syntax error near unexpected token 'done'

Question

I'm new to programming in bash and am trying to write a script. So far it's very rudimentary but I'm getting the above error with the done at the end.

for ((i = 1; i < 13; i++)) do
    if [ "$i" -lt "4" ]; then
        touch Block1/B8IT11"$i".txt
        echo B8IT11"$i" created
    else if [ "$i" -gt "3" -a  "$i" -lt "7" ]; then
        touch Block2/B8IT11"$i".txt
        echo B8IT11"$i" created
    else if [ "$i" -lt "6" -a  "$i" -lt "10" ]; then
        touch Block3/B8IT11"$i".txt
        echo B8IT11"$i" created
    else
        touch Block4/B8IT11"$i".txt
        echo B8IT11"$i" created
    fi
done

To my eyes I can't see the issue, as the if-else if-else ends with fi and the for loop should terminate with the done.

I've done cat -v and even dos2unix it. Does anyone see something I'm missing?

Try shellcheck.net.

chepner
– chepner

2016-07-02 13:23:27 +00:00
Commented Jul 2, 2016 at 13:23 — chepner
– chepner, Commented Jul 2, 2016 at 13:23

melpomene · Accepted Answer · 2016-07-02 13:00:24Z

1

There is no else if in bash. What you have is an else followed by a (nested) if construct. The outer else is unterminated (missing fi). Bash thinks you're still in an else block so it's not expecting done at this point:

for ((i = 1; i < 13; i++)) do
    if [ "$i" -lt "4" ]; then
        touch Block1/B8IT11"$i".txt
        echo B8IT11"$i" created
    else
        if [ "$i" -gt "3" -a  "$i" -lt "7" ]; then
            touch Block2/B8IT11"$i".txt
            echo B8IT11"$i" created
        else
            if [ "$i" -lt "6" -a  "$i" -lt "10" ]; then
                touch Block3/B8IT11"$i".txt
                echo B8IT11"$i" created
            else
                touch Block4/B8IT11"$i".txt
                echo B8IT11"$i" created
            fi
            done

Fix: Change all your else if to elif.

answered Jul 2, 2016 at 13:00

melpomene

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2 Comments

leviathan898 Over a year ago

Ah thanks, am coming from more of a c# background so completely missed that. Cheers!

chepner Over a year ago

There's also a missing semicolon before do.

sumitya · Accepted Answer · 2016-07-02 14:37:28Z

1

In bash else if is used as elif. you can try this:-

for ((i = 1; i < 13; i++)) do
    if [ "$i" -lt "4" ]; then
        touch Block1/B8IT11"$i".txt
        echo B8IT11"$i" created
    elif [ "$i" -gt "3" -a  "$i" -lt "7" ]; then
        touch Block2/B8IT11"$i".txt
        echo B8IT11"$i" created
    elif [ "$i" -lt "6" -a  "$i" -lt "10" ]; then
        touch Block3/B8IT11"$i".txt
        echo B8IT11"$i" created
    else
        touch Block4/B8IT11"$i".txt
        echo B8IT11"$i" created
    fi
done

answered Jul 2, 2016 at 14:37

sumitya

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