Bash – Alias or Intercept complex bash command with args

bash

How do you create a bash alias for a command with flags? ~~For example, if I run ls -l, I want that to alias to ls -a.~~

~~This doesn't work: alias "ls -l"="ls -a"~~

Another example: If I type reboot now, I want that to simply run reboot.

Would bash functions be useful here?

EDIT: Sorry for the silly ls -l example. This is what I really want: if you executed reboot now, I want it to actually execute reboot. This is why.

Best Answer

Another example: If I type reboot now, I want that to simply run reboot.

Create a function:

reboot() {
  command reboot
}

Now invoking the command reboot would invoke the function reboot which would ignore all parameters passed to it.

That is:

reboot

and

reboot now

would execute the command reboot (without any argument).

Related Solutions

Bash Alias – Create Dynamic Aliases Based on Location

You would be better served by a shell function:

doit () {
    local dir

    case $PWD/ in
        /home/alpha/*) dir=alpha ;;
        /home/beta/*)  dir=beta  ;;
        /home/gamma/*) dir=gamma ;;
        *) echo 'Not standing in the correct directory' >&2
           return 1
    esac

    python "/home/$dir/src/doit.py" --clean "$@"
}

This would set the variable dir to the string alpha, beta or gamma depending on the current working directory, or complain that you're in the wrong directory tree if the current directory is elsewhere.

It then runs the Python script, utilizing the $dir value, with the --clean option and adds whatever other arguments that you've passed to the function.

You would add this shell function's definition to wherever you ordinarily add aliases.

Bash Alias – Precedence and Shadowing Issues

Bash does allow aliases to contain aliases but it has built-in protections against infinite loops. In your case, when you type lsc, bash first expands the alias to:

ls -Flatr --color=always

Since ls is also an alias, bash expands it to:

lsc -Flatr --color=always

lsc is an alias but, quite sensibly, bash refuses to expand it a second time. If there was a program named lsc, bash would run it. But, there is not and that is why you get command not found.

Addendum

It is different when lscR runs. lscR expands to:

ls -FlatrR --color=always

Since ls is an alias, this expands to:

lsc -FlatrR --color=always

Since lsc is an alias, this expands to:

ls -Flatr --color=always -FlatrR --color=always

Since ls has already been expanded once, bash refuses to expand it a second time. Since a real command called ls exists, it is run.

History

As noted by Schily in the comments, bash borrowed the concept of not expanding an alias a second time from ksh.

Aside

Aliases are useful but not very powerful. If you are tempted to do something complex with an alias, such as argument substitution, don't; use a shell function instead.