how to loop through the content of a file using bash

Question

I want to print each line of a file using bash?

I'm using this script:

echo "Start!"
for p in (peptides.txt)
do
    echo "${p}"
done

and I get this output:

Start!
./runPep.sh: line 3: syntax error near unexpected token `('
./runPep.sh: line 3: `for p in (peptides.txt)'

(Later I want to do something more complicated with $p than just output to the screen.)

---

The environment variable SHELL is (from env):

SHELL=/bin/bash

/bin/bash --version output:

GNU bash, version 3.1.17(1)-release (x86_64-suse-linux-gnu)
Copyright (C) 2005 Free Software Foundation, Inc.

cat /proc/version output:

Linux version 2.6.18.2-34-default (geeko@buildhost) (gcc version 4.1.2 20061115 (prerelease) (SUSE Linux)) #1 SMP Mon Nov 27 11:46:27 UTC 2006

peptides.txt:

RKEKNVQ
IPKKLLQK
QYFHQLEKMNVK
IPKKLLQK
GDLSTALEVAIDCYEK
QYFHQLEKMNVKIPENIYR
RKEKNVQ
VLAKHGKLQDAIN
ILGFMK
LEDVALQILL

Answer 1

One of the ways to do it is:

while read p; do
  echo "$p"
done <peptides.txt

But this will result in whitespaces, interpretting, backlash sequences getting removed. If you have these concerns then you can try:

while IFS="" read -r p || [ -n "$p" ]
do
  printf '%s\n' "$p"
done < peptides.txt

---

Exceptionally, if the loop body may read from standard input, you can open the file using a different file descriptor:

while read -u 10 p; do
  ...
done 10<peptides.txt

10 is just random number here.

Answer 2

#!/bin/bash
for i in  `cat peptides.txt`
do
echo $i
done