help with choosing a file 1

[mzak94]

i need a command that lists the first few lines of files, i want to list the files first and then ask for a file to view

so far i have

#! /bin/bash

cd $1

ls

if [ "$(ls -A $1)" ]; then
read -p "show first few lines of files [y/n] " show
if [[ $show == [y] ]]; then
head -3 *
fi
else
echo "$1 is Empty"
fi

exit

can anyone helpme with my arguments

 

[PHV]

What is not working with your script ?
Anyway, I'd replace this:
if [ "$(ls -A $1)" ]; then
with this:
if [ "$(ls)" ]; then

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[KierenH]

Hi mzak94 I have recreated your script and it worked for me (on RHEL 5.3).

Are you getting any error messages that could help?

There is only one suggestion I would make in the script and one possible problem I can think of...

Suggestion: Instead of

if [ "$(ls -A $1)" ]; then

use

if [ "$(ls -A)" ]; then

as you have already "cd"'d to the directory you want to list.

Possible Problem: If you have too many files in the directory the "*" asterix will not be able to cope. As a solution to this you could use the find command with exec. i.e.:

find . -type f -maxdepth 1 -exec head -3 {} \;

but you wont get the pretty headers with file names. To get the headers you could do:

find . -type f -maxdepth 1 | awk '{printf "/n==> "$0" <==/n"; CMD="head -3 "$0; while ( CMD | getline OUT ) { print OUT } ; close(CMD)}'

but its a bit more complex a command

 

[PHV]

KierenH, a simpler way:

find . -type f -maxdepth 1 | xargs head -3

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886