argument not recognized for-loop

Question

I'm new pretty new when it comes to creating scripts in linux (I also don't really know on which terms to use like bash/shell), but I have the following command in a script which I want to execute with some arugments

write_loop.sh

#!/bin/bash
echo $1
echo $2
for i in {1..$1}; do printf "file '%s'\n" $2 >> list.txt; done

I'm trying to execute it like this in the terminal with 2 arguments: the amount of times to loop (150) and the filename to loop ("loop.mp4")

./write_loop.sh 150 loop.mp4

The arguments are passed to the script just fine, but for some reason the for-loop is not working. When I manually change $1 to 150 in write_loop.sh it does work, so I'm not sure what the issue is (maybe the argument is of a different type?). Any bit of help is appreciated.

Answer 1

Variables can't be used in range expansion {INT..INT}, so fallback to that is to use seq as show in related post. Alternatively, use C-like for loop since you're using bash anyway:

#!/bin/bash           
echo $1
echo $2
echo {1..$1}
for((i=1;i<=$1;i++)); do 
   printf "'%s'\n" "$2"
done

Or POSIX-ly:

#!/bin/bash           
echo $1
echo $2
i=1
while [ $i -le $1  ]; do 
   printf "'%s'\n" "$2"
   i=$((i+1))
done

Note I changed printf part for testing purposes. Adapt it to your needs.