Newline in bashscript variable

Question

I have the code:

for file in *
    do
      list="$list""$file "`cat $file | wc -l | sort -k1`$'\n'
   done

echo $list

This outputs a list like fileA 10 fileB 20 fileC 30.

I thought it should output:

fileA 10
fileB 20
fileC 30

I need the output to follow the latter as I then want to cycle through the $listvariable one line at a time and perform operations.

Could someone please help me out here?

Thank you.

Try `echo "$list"` instead. If you don't include the quotes the arguments will be split on whitespace and echo will separate them with a single space. — Gavin Smith, Aug 12 '14 at 09:12
possible duplicate of [Trying to embed newline in a variable in bash](http://stackoverflow.com/questions/9139401/trying-to-embed-newline-in-a-variable-in-bash) — tripleee, Aug 12 '14 at 10:28

score 1 · Accepted Answer · answered Aug 12 '14 at 09:14

1

it does have the new-line characters in the variable. when a variable is used outside of quotes it splits it into multiple arguments separated by whitespace.

what you want is echo "$list"

answered Aug 12 '14 at 09:14

programmerjake

1,794
11
15

score 0 · Answer 2 · answered Aug 12 '14 at 09:18

0

You can do it like this :

 for file in *
  do
     count=$(cat "$file" | wc -l)
     list="$list$file $count\n"
  done

echo -e $list

-e     enable interpretation of backslash escapes

answered Aug 12 '14 at 09:18

bachN

592
3
14

1

Or `wc -l <"$file"` to avoid that pesky UUOC. – tripleee Aug 12 '14 at 10:25
Exactly, `count=$(wc -l < "$file")` is better, +1 triplee – bachN Aug 12 '14 at 12:13

Newline in bashscript variable

2 Answers2