Bash Variables In For Loop Range
Answer : Yes, that's because brace-expansion occurs before parameter expansion. Either use another shell like zsh or ksh93 or use an alternative syntax: Standard (POSIX) sh syntax i=1 while [ "$i" -le "$number" ]; do echo "$i" i=$(($i + 1)) done Ksh-style for ((...)) for ((i=1;i<=number;i++)); do echo "$i" done use eval (not recommended) eval ' for i in {1..'"$number"'}; do echo "$i" done ' use the GNU seq command on systems where it's available 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). Avoid loops in shells. 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...