How can I prevent "type" stdout from appearing while running my script?

Question

I am currently working on a script that needs to check if a program is installed when it first starts up. If the program is not present, the script will proceed with doing whatever is necessary for the installation of the program. I have tried the following so far:

program_exist=$( type "someprogram" | grep "not found" )
if ! "$someprogram_exist"; then
        do some stuff
fi

program_exist=$( type "someprogram" 2>&1 >/dev/null | grep "not found" )
if ! "$someprogram_exist"; then
        do some stuff
fi

but each time I run this, I am always met with the following message:

./some_program.sh: line 8: ./some_program.sh: line 7: type: someprogram: not found: No such file or directory

Is there a way to check for the existence of a program without having it display the ./some_program.sh: line 8: ./some_program.sh: line 7: type: someprogram: not found: No such file or directory message each time?

see http://stackoverflow.com/questions/592620/how-to-check-if-a-program-exists-from-a-bash-script — Fredrik Pihl, Jan 12 '14 at 21:02

Gilles Quénot · Accepted Answer · 2014-01-12T21:13:09.580

3

Simply try this :

if ! type someprogram &>/dev/null; then
    do_some_stuff
fi

or shorter :

type someprogram &>/dev/null || do_some_stuff

edited Jan 12 '14 at 21:13

answered Jan 12 '14 at 21:07

Gilles Quénot

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thanks! looks like the way I had my 'if' statement setup was causing some issues, though it works like a charm now – lacrosse1991 Jan 12 '14 at 21:31

How can I prevent "type" stdout from appearing while running my script?

1 Answers1