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"$i" created
fi
done
To my eyes I can't see the issue, as the if-else if-else ends with fi and the for loop should terminate with the done.
I've done cat -v and even dos2unix it. Does anyone see something I'm missing?
There is no else if in bash. What you have is an else followed by a (nested) if construct. The outer else is unterminated (missing fi). Bash thinks you're still in an else block so it's not expecting done at this point:
for ((i = 1; i < 13; i++)) do
if [ "$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.
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