How to overwrite a bash command?

bash

Solution

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:

- be sure to reload your shell configuration: `source ~/.bashrc`

- 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.

Problem

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`?

Original source