Trying to alias a long command only gives me a partion of it

Question

I have the following command which works:

ls -la | awk '$5 > 2'

I'm trying to alias the entire thing. If I try a naive way:

alias ll "ls -la | awk '$5 > 2'"

It doesn't work.

If I try to escape the dollar sign, I get only a part of the command and ll is now ls -la | awk $5

What should I do in order to define this alias the way I want it to be?

Use a function instead of alias, then you don't have to worry about quoting issues. — Barmar, Aug 25 '16 at 19:53

Kusalananda · Answer 1 · 2017-09-22T17:15:23.650

1

Just use a shell function if your alias is more complicated than a simple command:

lz () {
    ls -la | awk '$5 > 2'
}

... or find a command that is simpler and fits as an alias:

alias lz='find . -maxdepth 1 -size +2c -ls'

The bash manual contains the statement

For almost every purpose, aliases are superseded by shell functions.

edited Sep 22 '17 at 17:15

answered Sep 12 '17 at 19:08

Kusalananda

score 0 · Answer 2 · answered Aug 25 '16 at 14:59

0

escape, equal

$ alias ll="ls -la | awk '\$5 > 2'"
$ alias|grep ll                                       
ll='ls -la | awk '\''$5 > 2'\'

answered Aug 25 '16 at 14:59

Ipor Sircer

That's not the resule i'm getting. The result i'm getting is ls -la | awk $5. – asaf92 Aug 25 '16 at 15:04
Which shell do you use? I've tried it on mksh & bash. – Ipor Sircer Aug 25 '16 at 15:21
@PanthersFan92: How do you check it? – Julie Pelletier Aug 25 '16 at 15:22
1

There is no need to grep an alias. This alias ll is faster and simpler. – Aug 26 '16 at 21:02

2 Answers2