Bash Variables In For Loop Range

Yes, that's because brace-expansion occurs before parameter expansion. Either use another shell like zsh or ksh93 or use an alternative syntax:

i=1 while [ "$i" -le "$number" ]; do   echo "$i"   i=$(($i + 1)) done

for ((i=1;i<=number;i++)); do   echo "$i" done

eval '   for i in {1..'"$number"'}; do     echo "$i"   done '

unset -v IFS # restore IFS to default for i in $(seq "$number"); do   echo "$i" done

(that one being less efficient as it forks and runs a new command and the shell has to reads its output from a pipe).

Using loops in a shell script are often an indication that you're not doing it right.

Most probably, your code can be written some other way.

You don't even need a for loop for this, just use the seq command:

$ seq 100

Here's the first 10 numbers being printed out:

$ seq 100 | head -10 1 2 3 4 5 6 7 8 9 10

You can use the following:

for (( num=1; num <= 100; num++ )) do     echo $num done

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