How can I use $variable in a shell brace expansion of a sequence?

Question

I want to use $var in a shell brace expansion with a range, in bash. Simply putting {$var1..$var2} doesn't work, so I went "lateral"...

The following works, but it's a bit kludgey.

# remove the split files
echo rm foo.{$ext0..$extN} rm-segments > rm-segments
source rm-segments

Is there a more "normal" way?

Answer 1

You may want to try :

eval rm foo.{$ext0..$extN}

Not sure whether this is the best answer, but it certainly is one.

Answer 2

As you already realized, {1..3} expands to 1 2 3 but {foo..bar} or {$foo..$bar} don't trigger brace expansion, and the latter is subsequently expanded to replace $foo and $bar by their values.

A fallback on GNU (e.g. non-embedded Linux) is the seq command.

for x in `seq $ext0 $extN`; do rm foo.$x; done

Another possibility. if foo. contains no shell special character, is

rm `seq $ext0 $extN | sed 's/^/foo./'`

The simplest solution is to use zsh, where rm foo.{$ext0..$extN} does what you want.

Answer 3

While the other answers discuss using eval and seq, in bash, you can use a traditional C style for loop in an arithmetic context. The variables ext0 and extN are expanded inside the ((..)) causing the loop to run for the range defined.

for (( idx = ext0; idx <= extN; idx++ )); do
    [[ -f "$foo.$idx" ]] || { printf "file %s does not exist" "$foo.$idx" >&2 ; continue ; }
    rm "$foo.$idx"
done

If you are looking for an optimal way and avoid multiple rm commands, you can use a temporary placeholder to store the filename results and call rm in one-shot.

 results=()
 for (( idx = ext0; idx <= extN; idx++ )); do
     [[ -f "$foo.$idx" ]] || { printf "file %s does not exist" "$foo.$idx" >&2 ; continue ; }
     results+=( "$foo.$idx" )
 done

and now call the rm command on the expanded array

 rm -- "${results[@]}"

Answer 4

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

    function f {
        echo $1
    }

    f_over_range 0 5
    f_over_range 00 05

Notes:

• Using eval exposes command injection security risk
• Linux will print "00\n01\n02..etc", but OSX will print "1\n2\n...etc"
• Using seq or C-style for loops won't match brace expansion's handling of leading zeros

Answer 5

I have the following command that finds the total number of files in my root directory:

command 01:

 ls -dq *dog* | wc -l

I also have a command to create multiple folders using a numeric sequence:

command 02:

mkdir -p {0..3}

I know in the root directory there are 3 files that have the string "DOG" in your name.

But I would like to use the 01 command in other root folders and that counted how many files have "DOG" in your name, store the result in variable result and used this variable in place of "3" as seen in command 02.

Below all my possible attempts without having been able to create subfolders 1, 2 and 3:

result=$ls -dq *dog* | wc -l) mkdir {0..$result}

result=$(ls -dq *dog* | wc -l) | mkdir {0..$result}

ls -dq *dog* | wc -l | xargs I{} mkdir {0..{}}





ls -dq *dog* | wc -l | while read file; do mkdir -p {1.."$file"}; done

But for any attempt above I only have two results: or is created a folder called {0..3} or nothing happens besides receiving a printing of the result of

 ls -dq *dog* | wc -l

Solved so bellow:

Looking at this answer it was possible to build a model capable of getting the desired result:

eval "mkdir -p {1..`ls -dq *dog* | wc -l`}"

While strictly is not the most appropriate answer because I had proposed that the problem was to "use a variable inside mkdir {1 .. $variable}"