How to skip comment line (#space) in bash for loop

Question

With the following code:

#!/bin/bash
export LC_ALL=C

for input_file in $(<inputflist.txt)
do
    case "$input_file" in \#*) continue ;; esac
    echo $input_file

done

And inputflist.txt

with the following content:

# foo.txt
bar.txt

I expect it to just print the last line bar.txt

, but instead print this:

foo.txt
bar.txt

What's the correct way to do this?

+3

neversaint 13 jul. 15 at 14:33

3 answers

while IFS= read -r line;do
[[ $line =~ ^[[:space:]]*# ]] || echo "$line"
done <file

Note:

+3

Jahid 13 jul. 15 at 14:45

what's wrong with

for input_file in $(grep -v "^#" inputflist.txt); do
  echo $input_file
done

or simply

grep -v "^#" inputflist.txt

?

+2

Maduraiveeran 13 jul. 15 at 15:12

anubhava · Accepted Answer · 2015-07-13T14:43:58+0000

This should work:

while IFS= read -r line; do
   [[ $line =~ ^[[:blank:]]*# ]] && continue
   echo "$line"
done < inputflist.txt

Output:

bar.txt