How to overwrite a bash command?

Question

So I basically am trying to overwrite my ssh command so I only have to type ssh and by default it would connect to my main server. Then if I passed it an argument, say username@server_port it would then run the basic command.

# Fast SSH  (a working progress) TODO: make work without naming the function `fssh`
function fssh() {

    ALEX_SERVER_CONNECTION=$ALEX_SERVER_UNAME@$ALEX_SERVER_PORT

    # if the `ssh` argument is not set
    if [ -z "${1+xxx}" ]; then
        # echo "ALEX_SERVER_CONNECTION is not set at all";
        ssh $ALEX_SERVER_CONNECTION
    fi

    # if the `ssh` argument is set
    if [ -z "$1" ] && [ "${1+xxx}" = "xxx" ]; then
        ssh $1
    fi
}

How do I get it to work without the f in front of the ssh?

So basically this is how it looks when properly done:

# Fast SSH
function ssh() {

    ALEX_SERVER_CONNECTION=$ALEX_SERVER_UNAME@$ALEX_SERVER_PORT

    # if the `ssh` argument is not set
    if [ -z "${1+xxx}" ]; then # ssh to the default server
        command ssh $ALEX_SERVER_CONNECTION
    fi

    # if the `ssh` argument is set
    if [ -z "$1" ] && [ "${1+xxx}" = "xxx" ]; then # ssh using a different server
        command ssh $1
    fi
}

Answer 1

Solution

You need to specify the absolute path to ssh command in your function otherwise it will be recursive. For instance, instead of:

function ssh() { ssh $USER@localhost; } # WRONG!

You should write:

function ssh() { command ssh $USER@localhost; }

Use command built-in to get the ssh from the PATH (as suggested by @chepner):

command [-pVv] command [arg ...]

  Run  command  with  args  suppressing  the normal shell function
  lookup. **Only builtin commands or commands found in the PATH  are
  executed**.

Why not Alias?

Using a function is the correct pattern, read the Aliases and Functions sections from the man page.

ALIASES

There is no mechanism for using arguments in the replacement text. If arguments are needed, a shell function should be used (see FUNCTIONS below).

Diagnostic

When naming your custom function ssh do the following:

1. be sure to reload your shell configuration: source ~/.bashrc
2. check what is ssh with: which ssh or type ssh

type and which

Prior to declaring a custom function I got:

type ssh  # → ssh is /usr/bin/ssh
which ssh # → /usr/bin/ssh

After declaring function ssh() { ssh my-vm; }, I got:

which ssh # → ssh () { ssh my-vm; }
type ssh  # → ssh is a shell function

Advices

Either use sh or bash syntax:

• sh: test is done with [ … ], portable but not powerful ;
• bash test is done [[ … ]] and function keyword, less portable but dev-friendly.

Answer 2

How do I get it to work without the f in front of the ssh?

Set an alias in ~/.bashrc:

alias ssh='fssh'