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)

Answer 1

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

Answer 2

Why not just this to print the first 50 files:

ls -1 | head -50

Answer 3

FILE="$(ls | head -1)"

Handled spaces in filenames correctly too when I tried it.

Answer 4

An addition to the answer of "Ayman Hourieh" and "Shawn Chin", in case it is needed for something else than content of a directory.

In newer version of bash you can use mapfile to store the directory in an array. See help mapfile

mapfile -t files_in_dir < <( ls )

If you want it completely in bash use printf "%s\n" * instead of ls, or just replace ls with any other command you need.

Now you can access the array as usual and get the data you need.

First element:

${files_in_dir[0]}

Last element (do not forget space after ":" ):

${files_in_dir[@]: -1}

Range e.g. from 10 to 20:

${files_in_dir[@]:10:20}

Attention for large directories, this is way more memory consuming than the other solutions.

Answer 5

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

Answer 6

My way would be:

ls | head -10 | tr "\n" " "

This will print the first 10 lines returned by ls, and then tr replaces all line breaks with spaces. Output will be on a single line.