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

Question

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 callsaction1` 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.

Solution 1

How about:

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

Solution 2

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

Solution 3

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

Solution 4

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.

View more solutions

Comments

[l0b0]

You might want to only include the second part of the answer - It's better than using eval and fits the question.

[nhed]

+1 IMO This is cleaner than having to eval as in the accepted answer, but the accepted answer is more specific to the OP

[Yogarine]

I would recommend against using eval and consider splicing up the arguments as suggested in Vijayender's answer.