How do you properly nest for loops inside if/else statements in shell script?

I have the following code in Linux:

#!/bin/sh
echo "Enter a number:"; read n;
if (($# != 0))
for ((i=1; i< $n+1; i++))
do
echo $i
done
else
for ((i=1; i<21; i++))
do
echo $i
done
fi


As you can tell, I am trying to print the values from 1 to n. If no user input is given, I automatically print from 1 to 20. When I run this script, it says I have syntax error near unexpected token else. Can somebody please help, I don't know what I'm missing.

else

for (( ... ))

/bin/sh

/bin/sh

/bin/bash

3 Answers
3

A version of your code that actually works with all /bin/sh implementations might look like:

/bin/sh

#!/bin/sh
echo "Enter a number:"; read n;

if [ "$#" -ne 0 ]; then
i=0; while [ "$i" -lt $(( n + 1 )) ]; do
echo "$i"
i=$((i + 1))
done
else
i=0; while [ "$i" -lt 21 ]; do
echo "$i"
i=$((i + 1))
done
fi


Note the then, needed for the if construct to be valid; the change from if (( ... )) to if [ ... ]; and the change to for ((;;;)) to a while loop for counting.

then

if

if (( ... ))

if [ ... ]

for ((;;;))

while

You are missing a then. In addition, as others have mentioned, your implementation is bashism, so note the change to the first line.

then

#!/bin/bash
echo "Enter a number:"; read n;
if (($# != 0))
then
for ((i=1; i< $n+1; i++))
do
echo $i
done
else
for ((i=1; i<21; i++))
do
echo $i
done
fi


for ((i=1; i<21; i++))

#!/bin/bash

You're missing then:

then

if (($# != 0)) ; then


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