Using command line argument range in bash for loop prints brackets containing the arguments

It's probably a lame question. But I am getting 3 arguments from command line [ bash script ]. Then I am trying to use these in a for loop.

for i in {$1..$2}
    do action1
done

This doesn't seem to work though and if $1 is "0" and $2 is 2 it prints {0..2}' and calls action1` only once. I referred to various examples and this appears to be the correct usage. Can someone please tell me what needs to be fixed here?

Thanks in advance.

You can slice the input using ${@:3} or ${@:3:8} and then loop over it

For eg., to print arguments starting from 3

for i in ${@:3} ; do echo $i; done

or to print 8 arguments starting from 3 (so, arguments 3 through 10)

for i in ${@:3:8} ; do echo $i; done

How about:

for i in $(eval echo {$1..$2}); do echo $i; done

Use the $@ variable?

for i in $@
do
    echo $i
done

If you just want to use 1st and 2nd argument , just

for i in $1 $2 

If your $1 and $2 are integers and you want to create a range, use the C for loop syntax (bash)

for ((i=$1;i<=$2;i++))
do
...
done

I had a similar problem. I think the issue is with dereferencing $1 within the braces '{}'. The following alternative worked for me ..

#!/bin/bash
for ((i=$1;i<=$2;i++))
do
   ...
done

Hope that helps.

for i in `seq $1 $2`; do echo $i; done

#/bin/bash
for i
do
  echo Value: $i
done

This will loop over all arguments given to the script file. Note, no "do" or anything else after the loop variable i.

I recommend this: In loop we handle only $1, and do shift N times (if arguments will be need next, then better save it, because shift removing argument):

for a in 1 2 3
do
echo $1
shift
done

Example: ./test.sh "abra k" "b 2" "c 14"

abra k
b 2
c 14