How do I check if two specific programs exist on a system in an if conditional?

Question

I have the following in my .bashrc to print a funny looking message:

fortune | cowsay -W 65

I don't want this line to run if the computer doesn't have fortune or cowsay installed.

What's the best or simplest way to perform this check?

Mihai Maruseac · Accepted Answer · 2013-01-24 06:29:38Z

1

You can use type or which or hash to test if a command exists.

From all of them, which works only with executables, we'll skip it.

Try something on the line of

if type fortune &> /dev/null; then
    if type cowsay &> /dev/null; then
        fortune | cowsay -W 65
    fi
fi

Or, without ifs:

type fortune &> /dev/null && type cowsay &> /dev/null && (fortune | cowsay -W 65)

answered Jan 24, 2013 at 6:29

Mihai Maruseac

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Comments

Community · Accepted Answer · 2017-05-23 10:34:57Z

1

type is the tool for this. It is a Bash builtin. It is not obsolete as I once thought, that is typeset. You can check both with one command

if type fortune cowsay
then
  fortune | cowsay -W 65
fi

Also it splits output between STDOUT and STDERR, so you can suppress success messages

type fortune cowsay >/dev/null
# or failure messages
type fortune cowsay 2>/dev/null
# or both
type fortune cowsay &>/dev/null

CommunityBot

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answered Jan 24, 2013 at 7:05

Zombo

1

Guru · Accepted Answer · 2013-01-24 06:27:20Z

0

If your intention is not to get the error messages appear in case of not being installed, you can make it like this:

(fortune | cowsay -W 65) 2>/dev/null

answered Jan 24, 2013 at 6:27

Guru

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