Question

I’m trying to do a seemingly easy thing: a script that takes a variable from the command line ($1) and uses this as the length of the for loop. So if the command is run.sh in terminal:

run.sh 5

will run the for loop 5 times. This is my attempt:

#!/bin/sh

for i in {1..$1};do
 echo $i
done

But the only output I receive is:

{1..5}

How can I solve this?

Note: I’m a total beginner in bash/shell scripting.

Answer 1

are you really using the bash shell. Not all shells (especially the true /bin/sh (bourne) ) support that feature. To disambiguate your situation, use

#!/bin/bash

at the top of your script.

One solution is to use

for i in $(eval echo  {1..$1} ); do echo $i; done

eval says, “for the current command string, in this case echo ${1..$1}, don’t execute the command yet, re-evaluate the complete command line for any expandable variables.”

If you system has the seq cmd, then you can rewrite this as

for i in $(seq 1 $1 ); do echo $i; done

seq is short for sequence, and can take any starting and ending number, and an optional ‘skip by’ factor, sa seq 0 5 100 will count by 5’s to 100.

I hope this helps.