Goal: using an input file with a list of file names, get the first 5 lines of each file and output to another file. Basically, I'm trying to find out what each program does by reading the header.

Shell: Ksh

Input: myfile.txt

tmp/file1.txt
tmp/file2.txt

Output:

tmp/file1.txt - "Creates web login screen"
tmp/file2.txt - "Updates user login"

I can use "head -5" but not sure how to get the input from the file. I'm assuming I could redirect (>> output.txt)the output for my output file.

Input file names use a relative path.

Update: I created a script below but I'm getting "syntax error: unexpected end of file". The script was created with VI.

! /bin/sh

cat $HOME/jmarti20.list | while read line
do
#echo $line" >> jmarti20.txt
head -n 5 /los_prod/$line >> $HOME/jmarti20.txt
done

Right, you can append output with >> to a file.

head -n 5 file1.txt >> file_descriptions.txt

You can also use sed to print lines, from documentation at pinfo sed.

sed 5q file1.txt >> file_descriptions.txt

Personal preference is to put file description in line 3, and only print line 3 of files.

sed -n 3p file1.txt >> file_descriptions.txt

The reasoning for using line 3 has to do with the first line often containing a "shebang" like #!/bin/bash, and the 2nd line having localization strings, such as # -*- coding: UTF-8 -*-, to allow proper display of extra character glyphs and languages in terminals and text editors that support them.

Below is what I came up with and seems to work fairly well:

#! /bin/sh
cat $HOME/jmarti20.list | while read line
do
 temp=$line
 temp2=$(head -n 5 /los_prod/$line)
 echo "$temp" "$temp2" >> jmarti20.txt
 #echo "$line" >> jmarti20.txt
 #head -n 5 /los_prod/$line >> $HOME/jmarti20.txt
done