How to cat multiple files from a list of files in Bash?

Question

I have a text file that holds a list of files. I want to cat their contents together. What is the best way to do this? I was doing something like this but it

Answer 1

I ended up with this:

sed '/^$/d;/^#/d;s/^/cat "/;s/$/";/' files | sh > output

Steps:

1. sed removes blank lines & commented out lines from files.

2. sed then wraps each line in cat " and ";.

3. Script is then piped to sh with and into output.

Answer 2

Or in a simple command

cat $(grep -v '^#' files) > output

Answer 3

xargs

The advantage of xargs over $(cat) is that cat expands to a huge list of arguments which could fail if you have a lot of files in the list:

printf 'a\nb\n#c\n' > files
printf '12\n3\n' > a
printf '4\n56\n' > b
printf  '8\n9\n' > c
# Optional grep to remove lines starting with #
# as requested by the OP.
grep -v '^#' files | xargs cat

Output:

12
3
4
56

Related: How to pipe list of files returned by find command to cat to view all the files

Answer 4

#!/bin/bash

files=()
while read; do
    case "$REPLY" in
        \#*|'') continue;;
        *) files+=( "$REPLY" );;
    esac
done < input
cat "${files[@]}"

What's better about this approach is that:

1. The only external command, cat, only gets executed once.
2. It's pretty careful to maintain significant whitespace for any given line/filename.

Answer 5

{
  while read file
  do
    #process comments here with continue
    cat "$file"
  done
} < tmp > newfile

Answer 6

How about cat $(cat listoffiles) | grep -v "^#"?