getting started:a simple ls program

rei · #1 (**permalink**) 08-22-04, 23:24

i'm just getting started with shell programming.

to start with, i tried to write a script which shows the result of ls.

Code:

#! /usr/local/bin/bash

echo "beginning of ls"

`ls -l `

echo "end of ls"

exit 0

It returns
beginning of ls
./test.sh: total: command not found
end of ls

It is because "total" was outputed from "ls -l"

Thus i changed my code to `ls -l` 2>/dev/null

Although i get no errors, i get no result of "ls -l" too..
just a simple of:
beginning of ls
./test.sh: total: command not found
end of ls

How can i execute "ls -l", at the same time displaying the result on screen, without asking the shell to execute lines displayed on screen (for example : total)

Thanks

rei · #2 (**permalink**) 08-22-04, 23:51

changed my code to the following

Code:

LIST=`ls -l`
echo $LIST

It works, but without newline.
Any other ways to implement it?

Thanks

SiverSloth · #3 (**permalink**) 08-23-04, 05:02

backticks are used when you want to put the output of a command into a variable. If you just want the output on stdout (the screen) use

Code:

#! /usr/local/bin/bash

echo "beginning of ls"

ls -l 

echo "end of ls"

exit 0

Damian Ibbotson · #4 (**permalink**) 08-23-04, 08:27

...and in your original 'echo $LIST' attempt, you could have quoted $LIST to retain the whitespace/newlines. e.g. echo "$LIST"

rei · #5 (**permalink**) 08-24-04, 01:58

Quote:

Originally Posted by Damian Ibbotson

...and in your original 'echo $LIST' attempt, you could have quoted $LIST to retain the whitespace/newlines. e.g. echo "$LIST"

Thanks very much damian!!!
I've grown up a bit from your guidance!

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