In bash, how can I print the first n elements of a list?

Question

In bash, how can I print the first n elements of a list?

For example, the first 10 files in this list:

FILES=$(ls)

UPDATE: I forgot to say that I want to print the elements on one line, just like when you print the whole list with echo $FILES.

Answer 1

FILES=(*)
echo "${FILES[@]:0:10}"

Should work correctly even if there are spaces in filenames.

FILES=$(ls) creates a string variable. FILES=(*) creates an array. See this page for more examples on using arrays in bash. (thanks lhunath)

Answer 2

FILES=$(ls)
echo $FILES | fmt -1 | head -10

Answer 3

echo $FILES | awk '{for (i = 1; i <= 10; i++) {print $i}}'

Edit: AAh, missed your comment that you needed them on one line...

echo $FILES | awk '{for (i = 1; i <= 10; i++) {printf "%s ", $i}}'

That one does that.

Answer 4

to do it interactively:

set $FILES && eval eval echo \\\${1..10}

to run it as a script, create foo.sh with contents

N=$1; shift; eval eval echo \\\${1..$N}

and run it as

bash foo.sh 10 $FILES