Bash: How to determine if a command exists without running it?

Question

I'm using Bash on OS X. There are several commands that I want to determine if exist, but don't want to run them in case they do exist and have unwanted effects. For example, if I wanted to test if destroyPC were a command, but didn't want to test it by running it. Is there a simple way to do this, like isCommand destroyPC?

EDIT: This is already answered here. I was looking for "does this COMMAND exist" without thinking to look for "does this PROGRAM exist". Remember that commands are just programs.

Check if `destroyPC` is in any of your `$PATH`. Or you should try `which destroyPC`. — Panwen Wang, Mar 10 '21 at 18:42
@PanwenWang Thanks, this does work, but the answer from the question Ostone0 linked is quicker. — gkeenley, Mar 10 '21 at 18:49
@Ostone0 Thanks, that does answer it. I updated the original question to redirect people to it. — gkeenley, Mar 10 '21 at 18:49

PowerBall253 · Answer 1 · 2021-03-10T19:53:35.980

0

Try using command -v <your-command>. It will return error code 1 if the command isn't found and 0 if it is, so you can use it like this:

if ! command -v foo &> /dev/null
    #command does not exist
else
    #command exists
fi

Here, I'm using the &> /dev/null to silence the output.

edited Mar 10 '21 at 19:53

answered Mar 10 '21 at 18:47

PowerBall253

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score 0 · Answer 2 · answered Mar 10 '21 at 18:50

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Interesting Question Lets say we want to test the existence of 'ls' commands, this is what I would type on the command line

command -v ls

If the commands doesnt exist, there will be no output If it does(assuming bash), the output will be /bin/ls

answered Mar 10 '21 at 18:50

Onke

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Bash: How to determine if a command exists without running it?

2 Answers2