How to execute a script only if it is present in bash?

Question

I wonder if there is an easier way to execute a script in bash only if this script exists. What I want is equivalent to:

if [ -x $name ]
then
  $name
fi

or

[ -x $name ] && $name

What I'm looking for is something like

exec_if_exist $name

which eliminates the repetition of the script name.

Is there a way to simplify this in bash? I do not need a function or "speculative" execution that would give the command an error not found.

The best

+6

Jaro Feb 21 '14 at 6:13

3 answers

Steven penny · Answer 1 · 2014-02-21T06:24:19+0000

You can simplify it with the command type. Your tests require a full or relative file path. With typeit, it will search for PATH.

type $name && $name

This is also good because it puts success on STDOUT and failure on STDERR, giving you full control over the output.

# mute success
type >/dev/null
# mute fail
type 2>/dev/null
# mute both
type &>/dev/null

nicky_zs · Answer 2 · 2014-02-21T06:37:07+0000

exec_if_exist() {
    test -x $1 && $1
}

, , $1.

serv-inc · Answer 3 · 2019-08-27T12:36:48+0000

type OS X. , ,

which -s command && command