[SOLVED] alias to list all the users

[searain]

$ gawk -F':' '{ print $1}' /etc/passwd

The above command will display all the users.

I make an alias for this command,

$ alias lsusers="gawk -F':' '{ print $1}' /etc/passwd"

The I use the alias to issue the command

$ lsusers

It is supposed to do be the same as

$ gawk -F':' '{ print $1}' /etc/passwd

But lsusers not just list users, it is the same as
cat /etc/passwd

What is wrong with this alias?

$ alias lsusers="gawk -F':' '{ print $1}' /etc/passwd"

[Irithori]

Hi and welcome

The problems are quoting and (lack) of escaping.
If you call alias after defining your lsusers, you will see this mess:

$ alias
alias lsusers='gawk -F'\'':'\'' '\''{ print }'\'' /etc/passwd'

As you can see, single quotes have been added and escaped. Which looks ugly.
But also the $1 didnt make it into the alias, as it has been already interpreted as empty by the shell.

So you could escape the $:

$alias lsuers="gawk -F':' '{ print \$1}' /etc/passwd"
$alias
alias lsusers='gawk -F'\'':'\'' '\''{ print $1}'\'' /etc/passwd'

And this actually works.

But this approach has issues:
- ugly quoting
- escapes needed
- hardcoded /etc/passwd
- limited to /etc/passwd
- required gawk, which may not be available


My suggestion is to use getent.
getent is part of the libc package, cut is in the core-utils package. Which means they *will* be available.
And getent will query the same user/group/network/etc backends (be it files or ldap or whatever) that the system itself uses.

So your alias would look like this:

$alias lsusers='getent passwd | cut -d ":" -f1'
$alias
alias lsusers='getent passwd | cut -d ":" -f1'